Advanced Statistics 201-337 • CEGEP, Montreal

Revision sheet: the continuous random variable and its density (201-337)

Q: Can a probability density exceed 1?

Yes, without any contradiction. What is capped at 1 is the total area under the curve, never the height. A uniform distribution on an interval of length 0.25 has a constant density equal to 4, and its area is indeed 0.25 times 4, which is 1. A density is a probability per unit: the more concentrated the variable, the taller it is.

Q: Why is the probability of an exact value zero?

Because a probability is an area, and the area above a single point, of zero width, is zero. That does not make the value impossible: the variable does take a value, and each one has probability zero. It is the reason why strict and non-strict inequalities give the same result in the continuous case, whereas they differ in the discrete case.

Q: How do you compute a continuous probability without integrals at CEGEP?

By computing a geometric area. A constant density gives a rectangle, a linear density a trapezoid or a difference of triangles, a step density a sum of rectangles. For the exponential distribution, the formula gives the answer directly, and for the normal distribution you read the table after computing the z-score. None of these moves requires integral calculus.

Q: What is the difference between a density and a cumulative distribution function?

The density is a height, the cumulative distribution function an accumulated area. The cumulative distribution function at x is the probability that the variable is less than or equal to x: it starts at 0, rises without ever coming back down and ends at 1. The density has no reason to do that and can go up, go down or exceed 1. The median is read from the cumulative distribution function, never from the density.

Q: What is the memoryless property of the exponential distribution?

It is the fact that a component already in service has exactly the same survival distribution as a new one. The probability of lasting five more years does not depend on the years already gone by: the failure rate is constant. The model suits an accidental electronic failure, but not a tire or a part that wears out, whose failure rate increases with age.

Q: How do you read the 68, 95 and 99.7 percent of the normal distribution?

They are areas, not heights, and they are the same for every normal distribution because they are measured in standard deviations. The band extending one standard deviation on each side of the mean carries 68.27 percent of the probability, the band at two standard deviations 95.45 percent, the one at three 99.73 percent. The rest is shared equally between the two tails.

This revision sheet covers the continuous random variable and its probability density function, part of the probability and statistics course 201-337 at CEGEP. It does not summarize the course: it answers a single question, what costs marks on the exam in this chapter, and which precise move avoids each loss.

The central move fits in three words, AREA VERSUS HEIGHT. The density $f(a)$ is not a probability and can exceed 1, the probability of an interval is the area above that interval, $P(X=a)=0$ for every value, and strict and non-strict inequalities give the same result. No integral calculus is used here: the areas are rectangles, triangles or trapezoids, and the normal distribution is read from the table or through its three bands.

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The thread of the chapter

A density is a HEIGHT, a probability is an AREA: on every question, say which area you are computing and on which figure, rectangle, triangle, trapezoid or band under the bell curve.

This chapter is part of 201-337, Advanced Statistics

What is on this sheet

1. The essentials
2. The rules in table form
3. The mistakes that cost marks
4. Which method to choose
5. How the answer is expected to be written
6. Check before you hand in
7. The typical problem, taken apart
8. Learn by heart
9. Frequently asked questions

Topics covered

Probability density function
Normalization condition
Probability as an area
Cumulative distribution function
Median of a continuous variable
Continuous uniform distribution
Exponential distribution
Memoryless property
Normal distribution
Empirical rule 68-95-99.7
Continuous expected value
Mode, median and mean

The essentials

The two conditions, and nothing else

• $f$ is a density if and only if TWO things are true: $f(x)\geq 0$ everywhere, and the total area under its curve is exactly 1.
• The second condition is Kolmogorov's axiom $P(\Omega)=1$ translated into an area. It is this condition, and never a value of $f$ , that fixes the constant of a density known only up to a factor.
• A probability is the AREA under $f$ above an interval: $P(a\leq X\leq b)$ is read between the curve, the horizontal axis and the vertical lines $x=a$ and $x=b$ .
• A height $f(a)$ is NOT a probability: it is a probability per unit, it carries a unit (per hour, per metre, per degree) and it can exceed 1.
• Immediate consequence: $P(X=a)=0$ for every value $a$ , because a segment of zero width has zero area. Hence $P(a\leq X\leq b)=P(a<X<b)$ , which is FALSE in the discrete case.

The three numbers 0.10, 0.25 and 0.15 are HEIGHTS, in probability per hour; the required probability is the shaded area,

1\times 0.10+1\times 0.25=0.35

Writing “the required probability is the area of the trapezoid with parallel sides $f(1)$ and $f(3)$ and width 2” before computing regularly earns half a mark for method, and above all keeps you from taking a triangle for a trapezoid.

The cumulative distribution function, the median and the expected value, without integrals

• $F(x)=P(X\leq x)$ is the area ACCUMULATED from the left. It increases from 0 to 1, it is continuous (no jump), and $P(a\leq X\leq b)=F(b)-F(a)$ .
• The median $m$ is the solution of $F(m)=0.5$ : it cuts the area into two halves, it does not cut the support in two.
• $E(X)$ is the $x$ -COORDINATE OF THE CENTRE OF GRAVITY of the area under $f$ . For a triangle, it is the mean of the $x$ -coordinates of its three vertices, which gives the expected value in one line.
• The peak of $f$ is the MODE, not the expected value. The three measures coincide only if the density is symmetric with a single peak, as for the uniform and the normal distributions.
• An approximate value of $E(X)$ is always obtained by cutting the support into strips and placing the weight of each strip at its centre: this is exactly the weighted mean of the discrete case.

For

f(x)=\dfrac{x}{8}

[0, 4]

, the median

\sqrt{8}\approx 2.83

cuts the AREA into two halves (blue, orange), not the support; the centre of gravity,

E(X)=\dfrac{8}{3}\approx 2.67

, is a little to the left.

No integral calculus is needed in this course. If the area is neither a rectangle, nor a triangle, nor a trapezoid, then the distribution is exponential or normal, and those two are read with a formula or from the table.

The three usual distributions, and how to take their area

• UNIFORM on $[a, b]$ : $f(x)=\dfrac{1}{b-a}$ , the area is a rectangle, so $P(c\leq X\leq d)=\dfrac{d-c}{b-a}$ , $E(X)=\dfrac{a+b}{2}$ and $V(X)=\dfrac{(b-a)^{2}}{12}$ .
• EXPONENTIAL with parameter $\lambda$ : $F(x)=1-e^{-\lambda x}$ and $P(X>x)=e^{-\lambda x}$ , so there is no area to cut up. $E(X)=\sigma=\dfrac{1}{\lambda}$ and median $\dfrac{\ln 2}{\lambda}$ , always smaller than the mean.
• NORMAL: maximum height $f(\mu)=\dfrac{1}{\sigma\sqrt{2\pi}}$ , which exceeds 1 as soon as $\sigma<0.399$ . The 68, 95 and 99.7 percent benchmarks are the AREAS of the bands $\mu\pm\sigma$ , $\mu\pm 2\sigma$ and $\mu\pm 3\sigma$ .
• Standardization $Z=\dfrac{X-\mu}{\sigma}$ divides widths by $\sigma$ and multiplies heights by $\sigma$ : the product, and therefore the area, does not move. That is why a single table serves every normal distribution.
• The exponential is MEMORYLESS, $P(X>s+t\mid X>s)=P(X>t)$ : the object does not age. The uniform, on the other hand, AGES: the waiting already done brings the bus closer.

Uniform: a rectangle; exponential: a curve decreasing from

\lambda

, area through

1-e^{-\lambda x}

; normal: a bell, area from the table.

The rules in table form

Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.

Discrete versus continuous, line by line

Half the marks lost in this chapter come from a reflex of the discrete case applied to the continuous case. The last column is the answer that holds HERE. A red cell means that the reading asked for does not exist and that you must switch to another object, not look harder.

The question	Discrete variable	Continuous variable
Describe the distribution	table of the $P(x_{i})$	density $f$ , area $=1$
Example: Die: six probabilities of $\frac{1}{6}$ . Steel bar: $f(x)=\frac{x}{8}$ on $[0, 4]$ , with area $\frac{1}{2}\times 4\times 0.5=1$ .
What you read at $a$	$P(X=a)$	$f(a)$ , per unit
Example: $f(4)=0.50$ per metre, which gives about $0.50\times 0.01=0.005$ over one centimetre.
$P(X=a)$	height of the bar	$0$
Example: $P(X=3)=\frac{1}{6}\approx 0.167$ for the die, against $P(X=3)=0$ on the bar.
Read $P$ at a POINT	possible	nothing to read reading impossible
Example: On the bar, $f(3)=0.375$ is not a probability; the area over $[2.95, 3.05]$ is $0.375\times 0.1=0.0375$ . What to do: Take an AREA over an interval of nonzero width, never a value at a point.
$\leq$ or $<$	different results	same result
Example: On the step density, $P(1\leq X\leq 3)=P(1<X<3)=0.35$ .
Maximum height	$\leq 1$	no ceiling
Example: Uniform on $[0, 0.25]$ : $f=4$ . Normal with standard deviation $0.25$ : $f(\mu)=1.5958$ .
The cumulative distribution function $F$	step function	continuous
Example: For the die, $F$ jumps by $\frac{1}{6}$ at 3. Here $F(x)=\frac{x^{2}}{16}$ , so $F(2)=0.25$ , without any jump.
$E(X)$	$\sum x_{i}P(x_{i})$	centre of gravity
Example: Triangle with vertices $(0, 0)$ , $(4, 0)$ and $(4, 0.5)$ : $E(X)=\frac{0+4+4}{3}\approx 2.67$ m.

Only one line of this table carries over from discrete to continuous unchanged, the one for the expected value: the mean is still weighted, with the density replacing the point probabilities.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Reading a density as a probability

2 marks, and the mistake repeats on every question of the chapter

What not to write

“ $f(4)=0.50$ , so there is a 50 percent chance that the flaw is at 4 metres.”

What to write

“ $f(4)=0.50$ is a density, in probability per metre. Over one centimetre around 4 metres, $P\approx 0.50\times 0.01=0.005$ .”

Why: The test that settles it in one second is the UNITS test: a probability has none, a density always has one. It is also the mistake that then leads to writing $P(X=a)=f(a)$ .

2. Believing that a density cannot exceed 1

1.5 marks, plus the time lost looking for an error that does not exist

What not to write

“I get $f(x)=4$ , that is impossible: a probability never exceeds 1, I must have made a mistake somewhere.”

What to write

“ $f(x)=4$ on $[0, 0.25]$ is correct: it is the AREA that equals 1, $0.25\times 4=1$ , and the height is not capped.”

Two uniform densities with the same area, 1. The one on the right is four times taller because it is four times narrower: a height of 2 breaks no rule.

Why: The more concentrated the variable, the taller the density: for a normal distribution, $\frac{1}{\sigma\sqrt{2\pi}}$ exceeds 1 as soon as $\sigma<0.399$ , which happens as soon as you measure in fine units.

3. Looking for the probability of an exact value on the curve

2 marks, and everything that follows rests on a number that does not exist

What not to write

“ $P(X=3)$ is read at the value 3: $P(X=3)=f(3)=0.375$ .”

What to write

“ $P(X=3)=0$ : the area above a point is zero. For a nonzero number, you need a width: $P(2.95\leq X\leq 3.05)\approx 0.375\times 0.1=0.0375$ .”

f(3)=0.375

is a HEIGHT (red); a probability needs a width: the band of width

0.1

around

3

(green) has area

0.0375

, and a single point,

0

Why: In the discrete case, $P(X=3)$ is the height of the jump of $F$ at 3. Here $F$ is CONTINUOUS, so it does not jump, so the probability of an isolated value is zero. The two worlds part ways exactly there.

4. Believing that a strict inequality changes the result

1 mark, and the result obtained is often absurd

What not to write

“ $P(1<X<3)=P(1\leq X\leq 3)-f(1)-f(3)=0.35-0.10-0.25=0$ .”

What to write

“ $P(1<X<3)=P(1\leq X\leq 3)=0.35$ : we remove two isolated values, so two ZERO areas.”

Why: The reflex comes from the discrete case, where the subtraction is right and equals exactly $P(X=1)+P(X=3)$ . In the continuous case these two terms are zero, and what gets subtracted by mistake are heights.

5. Fixing the constant with a value of the function instead of the area

the whole question, and the four parts that build on it

What not to write

“ $f(x)=kx$ on $[0, 4]$ , and it has to come to 1 at the end, so $f(4)=4k=1$ and $k=0.25$ .”

What to write

“The area equals 1: the curve bounds a triangle with base 4 and height $4k$ , so $\frac{1}{2}\times 4\times 4k=8k=1$ and $k=\frac{1}{8}=0.125$ .”

Requiring

f(4)=1

gives the red triangle, of area

2

; the condition bears on the area: height

0.5

, area

1

k=\dfrac{1}{8}

(green).

Why: No condition bears on the value of $f$ at a point: normalization is the SECOND condition of a density and it is written on the area. Here the error factor is exactly 2, that of the triangle.

6. Believing that the exponential distribution ages

2 marks on the interpretation question, which comes up on every exam

What not to write

“The component has already lasted 6 years, so it is closer to failing: its probability of lasting 5 more years is lower than that of a new one.”

What to write

“ $P(X>11\mid X>6)=\frac{e^{-2.2}}{e^{-1.2}}=e^{-1}=0.3679=P(X>5)$ : the exponential is memoryless, and the used component has the same survival distribution as a new one.”

The red curve, the survival curve of a component that has already lasted 6 years, is the curve of a new one simply shifted: same starting height, same decay.

Why: The density decreases, which suggests ageing, whereas it decreases only because fewer and fewer objects are still in service. The failure rate itself is CONSTANT. It is the UNIFORM distribution that ages, with the bus getting closer while you wait.

7. Confusing the density and the cumulative distribution function

2 marks, and the error spreads to the median and the quartiles

What not to write

“ $F(3)$ , I read it off the curve: $F(3)=0.375$ .”

What to write

“ $f(3)=0.375$ is the HEIGHT at 3; $F(3)=P(X\leq 3)$ is the area accumulated up to 3, namely $\frac{9}{16}=0.5625$ .”

Same steel bar, two curves:

f

rises to

0.5

then drops,

F

rises to

1

and stays there. Reading

F(3)

off the straight line gives

0.375

instead of

0.5625

Why: The two curves often live in the same window and look alike when $f$ is increasing. The sign that never misleads: $F$ starts at 0, rises without ever coming back down and ends at 1, whereas $f$ has no reason to do so.

8. Taking the peak of the density for the expected value

1.5 marks, and you lose the free visual check

What not to write

“The density is highest at $x=4$ , so the most probable value is 4 and $E(X)=4$ .”

What to write

“The maximum of $f$ is the MODE, here 4. The expected value is the centre of gravity of the area: $E(X)=\frac{0+4+4}{3}=\frac{8}{3}\approx 2.67$ , and the median is $\sqrt{8}\approx 2.83$ .”

The peak of

f

is at

4

: that is the mode (red). The expected value is the centre of gravity of the area,

\dfrac{0+4+4}{3}\approx 2.67

(green).

Why: Mode, median and mean coincide only if the density is symmetric with a single peak, which is the case for the uniform and the normal distributions: it is this special case that keeps the confusion alive.

9. Forgetting to split what lies outside a band in two

2 marks, and the announced risk is twice the real risk

What not to write

“The two-standard-deviation band carries $0.9545$ , so $P(X\geq 21.5)=1-0.9545=0.0455$ .”

What to write

“The remaining $0.0455$ is shared between the TWO tails, which are symmetric: $P(X\geq 21.5)=\frac{0.0455}{2}=0.0228$ .”

The central band, two standard deviations wide on each side, carries

0.9545

. The remaining

0.0455

is shared between the TWO tails, that is

0.0228

on each side.

Why: The complement of a CENTRED band is not one tail, it is two tails. The same division by two comes back in every calculation of the type $P(\left|X-\mu\right|\leq k)$ , and the table, which gives the area to the left, does not do it for you.

Which method to choose

Which area, and on which figure

Look at the SHAPE of the density, not the topic of the statement. Each shape has its figure, and each figure has its area.

The figures the tree chooses between: a rectangle (uniform), a triangle or a trapezoid (linear density), a curve whose area is read with a formula or from the table.

If CONSTANT density on $[a, b]$ (uniform distribution) → rectangle: the probability is the ratio of lengths, $\dfrac{d-c}{b-a}$

Example: $P(3\leq X\leq 7)$ on $[0, 12]$ equals $\frac{4}{12}=\frac{1}{3}\approx 0.3333$

it is the only distribution of the chapter where a probability is a proportion of length
If LINEAR density, of the type $f(x)=kx$ or $f(x)=k(b-x)$ → trapezoid with parallel sides $f(c)$ and $f(d)$ , or a difference of two triangles

Example: $f(x)=\frac{x}{8}$ : $P(1\leq X\leq 3)=\frac{0.125+0.375}{2}\times 2=0.50$

both routes give the same number, since a trapezoid is a difference of triangles
If STEP density and an interval straddling two steps → cut into rectangles, one per step, then add up the areas

Example: $P(1\leq X\leq 3)=1\times 0.10+1\times 0.25=0.35$

you add AREAS, never heights
If density $f(x)=\lambda e^{-\lambda x}$ (exponential distribution) → no area to cut up: $P(X>x)=e^{-\lambda x}$ and $F(x)=1-e^{-\lambda x}$

Example: $\lambda=0.2$ : $P(X>8)=e^{-1.6}\approx 0.2019$

keeping the form $P(X>x)=e^{-\lambda x}$ saves a complement on every line
If BELL-shaped density (normal distribution) → either the table after $z=\dfrac{x-\mu}{\sigma}$ , or the 68, 95 and 99.7 percent bands

Example: $\mu=21$ , $\sigma=0.25$ : $z=\frac{21.5-21}{0.25}=2$ so $P(X\geq 21.5)=0.0228$

the band avoids the table when the endpoint falls exactly on a whole number of standard deviations
If the statement asks for a CUTOFF value, not a probability → work back through the cumulative distribution function: solve $F(t)=p$ , with the logarithm for the exponential

Example: $1-e^{-0.25t}=0.90$ gives $t=\frac{\ln 10}{0.25}\approx 9.21$ minutes

the median is the case $p=0.5$ , the quartiles the cases $0.25$ and $0.75$

No branch calls for an integral, and that is deliberate: this course assumes none. If the area you are looking for is neither a rectangle, nor a triangle, nor a trapezoid, then the distribution is exponential or normal, and those two are read with a formula or from the table.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Showing that a function is a density and finding the constant

When to use it: The statement gives $f(x)=kx$ , $f(x)=k(b-x)$ or a step function, and asks for $k$ , or asks you to check that $f$ is indeed a density.

1 State the TWO conditions, by name: nonnegativity of $f$ everywhere, and total area equal to 1.
2 Check nonnegativity by giving the reason, not by asserting it: “on $[0, 4]$ , $x\geq 0$ , so $kx\geq 0$ as long as $k>0$ ; elsewhere $f$ is zero”.
3 Name the FIGURE whose area you take before computing: “the curve bounds a right triangle with base 4 and height $4k$ ”.
4 Write the area equation and solve it: $\frac{1}{2}\times 4\times 4k=8k=1$ , so $k=\frac{1}{8}=0.125$ .
5 Write $f$ out in full, piece by piece, zero included: $f(x)=\frac{x}{8}$ on $[0, 4]$ and $f(x)=0$ elsewhere.

Concluding sentence

“ $f$ is nonnegative on $\mathbb{R}$ and the area under its curve is $8k$ . The normalization condition requires $8k=1$ , so $f$ is a probability density function if and only if $k=\frac{1}{8}$ , with $f(x)=\frac{x}{8}$ on $[0, 4]$ and $f(x)=0$ elsewhere.”

The trap: Forgetting the last line, the one that says $f$ is zero outside the support. Without it, the total area is not the one you have just computed and the proof is incomplete.

Marking: 1 mark for the justified nonnegativity, 2 marks for the area equation and its solution, 1 mark for writing $f$ out in full.

Computing a probability by naming the area

When to use it: Any question that starts with “compute $P(\dots)$ ” on a continuous variable, including when the event is written with an absolute value or with the words “at least”.

1 Translate the event into an INTERVAL: $\left|E\right|\leq 0.2$ becomes $[-0.2, 0.2]$ , “at least 3 hours” becomes $[3, +\infty)$ .
2 Say which figure carries this area: rectangle, trapezoid, difference of triangles, or band under the bell curve.
3 Compute the area with both dimensions written out in full, width then height, and never a height alone.
4 Check that the result falls between 0 and 1, and that it is smaller than the probability of an interval that contains it.

Concluding sentence

“The event $1\leq X\leq 3$ corresponds to the area under $f$ between the vertical lines $x=1$ and $x=3$ . This area splits into two rectangles: $P(1\leq X\leq 3)=1\times 0.10+1\times 0.25=0.35$ .”

The trap: Answering with the height when the interval is narrow. A width of $0.01$ does not exempt you from multiplying: $f(21.0)\times 0.01=1.5958\times 0.01=0.0160$ , and it is this last number that is the probability.

Marking: 0.5 mark for the interval written down, 0.5 for the name of the figure, 1 mark for the area calculation.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The total area equals 1, always

Before the first probability, recompute the total area under the density, figure by figure. It is the only check that validates both the constant and the way you cut the region.

Three rectangles of $0.20$ , $0.50$ and $0.30$ : the sum is $1.00$ , all is well. A sum of $0.50$ signals that you added the HEIGHTS $0.10+0.25+0.15$ instead of the areas.
The probability stays between 0 and 1, the density does not

Apply the bound to the right object. A probability greater than 1 stops the calculation at once; a density of 4 or of 1.6 proves nothing against you.

$P(X>8)=e^{-1.6}=0.2019$ passes. If your calculator shows $4.95$ , it is the minus sign of the exponent that got dropped, never a property of the distribution.
The complement adds back up to 1

Cut the support into disjoint pieces and add them up: the sum must be exactly 1. This catches the forgotten area of a step and the complement taken the wrong way round.

$0.4512+0.3469+0.2019=1.0000$ over the three pieces of the support. A sum of $0.98$ says that a piece is missing, not that you are allowed to round.
The expected value falls on the heavy side of the density

Look at where the density is high: $E(X)$ is the balance point of the area, so it leans to that side. It is free and it catches factor errors.

Increasing density on $[0, 4]$ : $E(X)=2.67$ is indeed to the right of the midpoint 2. An answer of $1.5$ is impossible before you even redo the calculation.
The median comes from $F$ , not from the midpoint of the support

Solve $F(m)=0.5$ , then compare $m$ with the midpoint of the support: the median leans towards the side where the density is high, like the expected value.

$\frac{m^{2}}{16}=0.5$ gives $m=\sqrt{8}\approx 2.83$ on $[0, 4]$ , and not 2. For an exponential, the median $\frac{\ln 2}{\lambda}$ is always SMALLER than the mean $\frac{1}{\lambda}$ .

The typical problem, taken apart

A linear density on a narrow support, from the constant to the centre of gravity

A workshop measures the EXCESS thickness $X$ , in millimetres, of a coat of paint. This excess never exceeds $0.5$ mm, and the larger an excess, the more frequent it is: the density is therefore modelled by $f(x)=kx$ for $0\leq x\leq 0.5$ , and $f(x)=0$ elsewhere.

Find $k$ , comment on the value of $f(0.5)$ , write $F$ , then compute $P(X\geq 0.3)$ , the median, the expected value, the variance, and finally $P(X=0.3)$ .

We take as given that the variance of a triangular density on $[a, b]$ with peak $c$ is $V(X)=\dfrac{a^{2}+b^{2}+c^{2}-ab-ac-bc}{18}$ .

The density is zero outside

[0, 0.5]

and rises in a straight line from the origin: the area under it is a triangle with base

0.5

and height

f(0.5)=0.5k

Step 1

Two conditions. NONNEGATIVITY: on $[0, 0.5]$ , $x\geq 0$ so $kx\geq 0$ as long as $k>0$ , and $f$ is zero elsewhere. AREA: the curve bounds a right triangle with base $0.5$ and height $0.5k$ , of area $\frac{1}{2}\times 0.5\times 0.5k=0.125k$ . Solve $0.125k=1$ , so $k=8$ and $f(x)=8x$ on $[0, 0.5]$ .

Why

The equation that gives the constant is ALWAYS the area equation, never a value of $f$ . Naming the triangle before computing earns half a mark for method and keeps you from confusing it with a trapezoid.

Step 2

$f(0.5)=8\times 0.5=4$ , in reciprocal millimetres. The density exceeds 1, and yet the area is $\frac{1}{2}\times 0.5\times 4=1$ .

Why

This is the check that reassures instead of worrying. The support is only half a millimetre long, so the probability PER millimetre is bound to be large: a tall height and an area equal to 1 do not contradict each other.

Step 3

$F(x)=P(X\leq x)$ is the area of the triangle with base $x$ and height $f(x)=8x$ , namely $F(x)=\frac{1}{2}\times x\times 8x=4x^{2}$ on $[0, 0.5]$ , with $F(x)=0$ before 0 and $F(x)=1$ after $0.5$ . Check: $F(0.5)=4\times 0.25=1$ .

Why

A cumulative distribution function that does not end at 1 is wrong, and this check takes two seconds. In particular it catches the factor $\frac{1}{2}$ of the triangle, which is the most frequent omission at this step.

Step 4

$P(X\geq 0.3)=1-F(0.3)=1-4\times 0.09=1-0.36=0.64$ .

Why

The complement saves a trapezoid and an error in the endpoints. The result is plausible: the density increases, so the right half of the support carries most of the probability.

Step 5

Median: $F(m)=0.5$ , so $4m^{2}=0.5$ , so $m^{2}=0.125$ and $m=\sqrt{0.125}\approx 0.354$ mm.

Why

The median is read from $F$ and never from $f$ . Immediate check: it falls to the RIGHT of the midpoint $0.25$ , as an increasing density requires.

Step 6

Expected value: the area under $f$ is the triangle with vertices $(0, 0)$ , $(0.5, 0)$ and $(0.5, 4)$ . The $x$ -coordinate of its centre of gravity is the mean of the $x$ -coordinates of the three vertices, $\frac{0+0.5+0.5}{3}=\frac{1}{3}\approx 0.333$ mm.

Why

No integral: the centre of gravity of a triangle is the mean of its three vertices, and that is exactly the geometric definition of the continuous expected value. One line replaces a calculation that this course does not ask for.

Step 7

Variance, with $a=0$ , $b=0.5$ and a peak at $c=0.5$ : $V(X)=\frac{0+0.25+0.25-0-0-0.25}{18}=\frac{0.25}{18}\approx 0.013889$ and $\sigma=\sqrt{0.013889}\approx 0.1179$ mm.

Why

The standard deviation must stay small compared with the width of the support, here $0.12$ against $0.5$ . A standard deviation larger than the whole support would signal a formula error, not a very spread-out variable.

Step 8

$P(X=0.3)=0$ , whereas $f(0.3)=8\times 0.3=2.4$ . To get a nonzero number, you need a width: $P(0.29\leq X\leq 0.31)=\frac{f(0.29)+f(0.31)}{2}\times 0.02=2.4\times 0.02=0.048$ .

Why

This is the question that separates the papers. The two statements do not contradict each other: one is about an area of zero width, the other about a height. And here the trapezoid gives exactly $F(0.31)-F(0.29)$ .

Step 9

Final check: $E(X)\approx 0.333$ , $m\approx 0.354$ and the mode is $0.5$ , so mean, then median, then mode. And $F(0.5)=1$ .

Why

This order is the signature of a density skewed to the LEFT, which the sketch confirms. If it came out reversed, the error would be in the centre of gravity, not in the median, which has been checked on $F$ .

The conclusion, written out

The excess thickness has density $f(x)=8x$ on $[0, 0.5]$ and cumulative distribution function $F(x)=4x^{2}$ . Its median is $\sqrt{0.125}\approx 0.354$ mm, its expected value $\frac{1}{3}\approx 0.333$ mm and its standard deviation $\approx 0.118$ mm; $64$ percent of the parts exceed $0.3$ mm of excess, and the probability of an excess EXACTLY equal to $0.3$ mm is zero.

The classic mistake on this problem: Writing $f(0.5)=1$ “so that it comes to 1”, hence $k=2$ . Everything that follows stays consistent with itself, $F$ , the median and the expected value come out without any warning, and only the check $F(0.5)=1$ catches the mistake: it would give $0.25$ .

Learn by heart

• Two conditions, never one more: $f(x)\geq 0$ EVERYWHERE, and total area equal to 1. It is the second one that gives the constant.
• A probability is an AREA and has no unit; a density is a HEIGHT and always has one. It can exceed 1.
• $P(X=a)=0$ for every value, so $P(a\leq X\leq b)=P(a<X<b)$ . True for continuous, FALSE for discrete.
• $F(x)=P(X\leq x)$ increases from 0 to 1 WITHOUT JUMPS; $P(a\leq X\leq b)=F(b)-F(a)$ and the median solves $F(m)=0.5$ .
• Uniform on $[a, b]$ : $f=\frac{1}{b-a}$ , $E=\frac{a+b}{2}$ , $V=\frac{(b-a)^{2}}{12}$ . It AGES.
• Exponential: $F(x)=1-e^{-\lambda x}$ , $P(X>x)=e^{-\lambda x}$ , $E=\sigma=\frac{1}{\lambda}$ , median $\frac{\ln 2}{\lambda}$ . It is MEMORYLESS.
• Normal: $f(\mu)=\frac{1}{\sigma\sqrt{2\pi}}$ exceeds 1 as soon as $\sigma<0.399$ , and 68, 95 then 99.7 percent are the AREAS of the bands $\mu\pm\sigma$ , $\mu\pm 2\sigma$ and $\mu\pm 3\sigma$ .
• $E(X)$ is the CENTRE OF GRAVITY of the area under $f$ ; the peak of $f$ is the MODE, and the two coincide only for a symmetric density.

Frequently asked questions

Can a probability density exceed 1?

Yes, without any contradiction. What is capped at 1 is the total area under the curve, never the height. A uniform distribution on an interval of length 0.25 has a constant density equal to 4, and its area is indeed 0.25 times 4, which is 1. A density is a probability per unit: the more concentrated the variable, the taller it is.

Why is the probability of an exact value zero?

Because a probability is an area, and the area above a single point, of zero width, is zero. That does not make the value impossible: the variable does take a value, and each one has probability zero. It is the reason why strict and non-strict inequalities give the same result in the continuous case, whereas they differ in the discrete case.

How do you compute a continuous probability without integrals at CEGEP?

By computing a geometric area. A constant density gives a rectangle, a linear density a trapezoid or a difference of triangles, a step density a sum of rectangles. For the exponential distribution, the formula gives the answer directly, and for the normal distribution you read the table after computing the z-score. None of these moves requires integral calculus.

What is the difference between a density and a cumulative distribution function?

The density is a height, the cumulative distribution function an accumulated area. The cumulative distribution function at x is the probability that the variable is less than or equal to x: it starts at 0, rises without ever coming back down and ends at 1. The density has no reason to do that and can go up, go down or exceed 1. The median is read from the cumulative distribution function, never from the density.

What is the memoryless property of the exponential distribution?

It is the fact that a component already in service has exactly the same survival distribution as a new one. The probability of lasting five more years does not depend on the years already gone by: the failure rate is constant. The model suits an accidental electronic failure, but not a tire or a part that wears out, whose failure rate increases with age.

How do you read the 68, 95 and 99.7 percent of the normal distribution?

They are areas, not heights, and they are the same for every normal distribution because they are measured in standard deviations. The band extending one standard deviation on each side of the mean carries 68.27 percent of the probability, the band at two standard deviations 95.45 percent, the one at three 99.73 percent. The rest is shared equally between the two tails.

Practise it

Corrected exercises: Continuous random variables and density, 201-337

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

11 corrected exercises
110 points
150 minutes

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Revision sheet: the continuous random variable and its density (201-337)

The essentials

The two conditions, and nothing else

The cumulative distribution function, the median and the expected value, without integrals

The three usual distributions, and how to take their area

The rules in table form

Discrete versus continuous, line by line

The mistakes that cost marks

1. Reading a density as a probability

2. Believing that a density cannot exceed 1

3. Looking for the probability of an exact value on the curve

4. Believing that a strict inequality changes the result

5. Fixing the constant with a value of the function instead of the area

6. Believing that the exponential distribution ages

7. Confusing the density and the cumulative distribution function

8. Taking the peak of the density for the expected value

9. Forgetting to split what lies outside a band in two

Which method to choose

Which area, and on which figure

How the answer is expected to be written

Showing that a function is a density and finding the constant

Computing a probability by naming the area

Check before you hand in

The typical problem, taken apart

A linear density on a narrow support, from the constant to the centre of gravity

Learn by heart

Frequently asked questions

Can a probability density exceed 1?

Why is the probability of an exact value zero?

How do you compute a continuous probability without integrals at CEGEP?

What is the difference between a density and a cumulative distribution function?

What is the memoryless property of the exponential distribution?

How do you read the 68, 95 and 99.7 percent of the normal distribution?

Corrected exercises: Continuous random variables and density, 201-337

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