IB Mathematics Analysis and Approaches Topic AHL…

Discrete And Continuous Random Variables: Spread, Densities And Transformations

In 4.7 you found the expected value of a discrete random variable, the long-run average outcome, but the average is only half the story.
Two lottery games can share the same expected payout while one is steady and the other wildly unpredictable, and to capture that difference you need a measure of spread, the variance.
This subtopic does three connected jobs: it measures spread for discrete variables, extends the toolkit to continuous variables through a density function, and shows how a linear transformation shifts and scales both.

Variance Of A Discrete Random Variable

Variance is the average squared distance of the outcomes from the mean $\mu = E(X)$, and squaring keeps the below-mean and above-mean gaps from cancelling.
In practice you almost always use the shortcut form below. $$\operatorname{Var}(X) = E(X^{2}) - [E(X)]^{2}$$
1. Find the mean $E(X) = \sum x\,P(X = x)$.
2. Find $E(X^{2}) = \sum x^{2}\,P(X = x)$, weighting the squared values by the same probabilities.
3. Subtract them, then take the square root for the standard deviation.

Note

The formula booklet gives $\operatorname{Var}(X) = E(X^{2}) - [E(X)]^{2}$ under Topic 4 (AHL).
The standard deviation is $\sigma = \sqrt{\operatorname{Var}(X)}$, back in the original units of $X$.

Example question

The discrete random variable $X$ has the probability distribution below.

x	1	2	3
P(X = x)	0.2	0.5	0.3

Find $E(X)$, $\operatorname{Var}(X)$ and the standard deviation of $X$.

Solution

$$E(X) = 1(0.2) + 2(0.5) + 3(0.3) = 2.1$$

$$E(X^{2}) = 1^{2}(0.2) + 2^{2}(0.5) + 3^{2}(0.3) = 0.2 + 2.0 + 2.7 = 4.9$$

$$\operatorname{Var}(X) = 4.9 - 2.1^{2} = 4.9 - 4.41 = 0.49$$

$$\sigma = \sqrt{0.49} = 0.7$$

From Sums To Integrals: The Probability Density Function

A continuous random variable can take any value in a range, like a waiting time or a length, so there are infinitely many possible values and you cannot list a probability for each one.
Instead a probability density function $f(x)$ describes how probability is spread out, and probability becomes area under the curve. $$P(a \le X \le b) = \int_{a}^{b} f(x)\,dx$$
1. $f(x) \ge 0$ everywhere, since area cannot be negative.
2. $\displaystyle\int_{-\infty}^{\infty} f(x)\,dx = 1$, since the total probability is 1.

Common Mistake

A density is not a probability, so $f(x)$ can be greater than 1; only the area under it is a probability.
The probability of any single exact value is an area of zero width, so $P(X = c) = 0$, and $P(X < c)$ equals $P(X \le c)$.
So strict and non-strict inequalities give the same probability for continuous variables.

Mean, Mode, Median And Variance For Continuous Variables

Every discrete formula carries over by replacing the sum $\sum$ with an integral $\int$ and the probability $P(X=x)$ with $f(x)\,dx$. $$E(X) = \int x\,f(x)\,dx, \qquad E(X^{2}) = \int x^{2} f(x)\,dx, \qquad \operatorname{Var}(X) = E(X^{2}) - [E(X)]^{2}$$
The mode: The value where $f(x)$ is greatest, the peak of the density, found by maximising $f$ and checking endpoints too.
The median $m$: The value that splits the area in half, so $\displaystyle\int_{-\infty}^{m} f(x)\,dx = \tfrac{1}{2}$.

Note

The formula booklet gives the continuous $E(X) = \int x f(x)\,dx$ and the variance, so those are provided.
The mode and median come from the density itself, found by maximising $f$ and by splitting the area, not from a booklet formula.

Example question

A continuous random variable $X$ has probability density function $f(x) = kx$ for $0 \le x \le 2$, and $f(x) = 0$ otherwise.

(a) Find $k$.

(b) Find the mean.

(d) State the mode.

(e) Find the variance.

Solution

(a): total area is 1, so $\int_{0}^{2} kx\,dx = k\left[\tfrac{x^{2}}{2}\right]_{0}^{2} = 2k = 1$.
$k = \tfrac{1}{2}$, so $f(x) = \tfrac{x}{2}$.
(b): $E(X) = \int_{0}^{2} x \cdot \tfrac{x}{2}\,dx = \left[\tfrac{x^{3}}{6}\right]_{0}^{2} = \tfrac{8}{6} = \tfrac{4}{3} \approx 1.33$.
(c): median $m$ solves $\int_{0}^{m} \tfrac{x}{2}\,dx = \tfrac{m^{2}}{4} = \tfrac{1}{2}$, so $m = \sqrt{2} \approx 1.41$.
(d): $f(x) = \tfrac{x}{2}$ increases across $[0, 2]$, so the mode is at the endpoint $x = 2$.
(e): $E(X^{2}) = \int_{0}^{2} x^{2}\cdot\tfrac{x}{2}\,dx = \left[\tfrac{x^{4}}{8}\right]_{0}^{2} = 2$, so $\operatorname{Var}(X) = 2 - \left(\tfrac{4}{3}\right)^{2} = \tfrac{2}{9} \approx 0.222$.

Linear Transformations: Shifting And Scaling A Variable

Changing units or adding a constant turns $X$ into $aX + b$, like converting Celsius to Fahrenheit or adding a fixed bonus to every score.
The mean and variance respond in a predictable way that works for both discrete and continuous variables. $$E(aX + b) = a\,E(X) + b \qquad\qquad \operatorname{Var}(aX + b) = a^{2}\operatorname{Var}(X)$$
1. Adding $b$ slides every outcome along by $b$, so the mean shifts by $b$ but the spread is unchanged.
2. Multiplying by $a$ scales the mean by $a$ and the variance by $a^{2}$, because variance is built from squared distances.
3. The standard deviation therefore scales by $|a|$, and the constant $b$ has no effect on it at all.

Exam technique

These two transformation results are not in the formula booklet, so learn them.
The two traps are forgetting to square the $a$ in the variance, and wrongly letting $b$ change the variance.

Example question

The variable $X$ from the earlier table has $E(X) = 2.1$ and $\operatorname{Var}(X) = 0.49$. Let $Y = 3X + 5$. Find $E(Y)$, $\operatorname{Var}(Y)$ and the standard deviation of $Y$.

Solution

$$E(Y) = 3E(X) + 5 = 3(2.1) + 5 = 11.3$$ $$\operatorname{Var}(Y) = 3^{2}\operatorname{Var}(X) = 9(0.49) = 4.41$$

$$\sigma_{Y} = \sqrt{4.41} = 2.1$$ which is 3 times $\sigma_{X}$, confirming the standard deviation scales by $|a|$.

Active recall

Write down the shortcut formula for the variance of a random variable in terms of $E(X^{2})$ and $E(X)$.
For a continuous random variable, what do you integrate to find a probability, and why is $P(X = c) = 0$?
How do you locate the median $m$ of a continuous random variable from its density $f(x)$?
State how $E(X)$ and $\operatorname{Var}(X)$ change under the transformation $aX + b$, and explain why $b$ does not affect the variance.
Give the two conditions a function must satisfy to be a valid probability density function.

Discrete And Continuous Random Variables: Spread, Densities And Transformations

In 4.7 you found the expected value of a discrete random variable, the long-run average outcome, but the average is only half the story.
Two lottery games can share the same expected payout while one is steady and the other wildly unpredictable, and to capture that difference you need a measure of spread, the variance.
This subtopic does three connected jobs: it measures spread for discrete variables, extends the toolkit to continuous variables through a density function, and shows how a linear transformation shifts and scales both.

Variance Of A Discrete Random Variable

Variance is the average squared distance of the outcomes from the mean $\mu = E(X)$, and squaring keeps the below-mean and above-mean gaps from cancelling.
In practice you almost always use the shortcut form below. $$\operatorname{Var}(X) = E(X^{2}) - [E(X)]^{2}$$
1. Find the mean $E(X) = \sum x\,P(X = x)$.
2. Find $E(X^{2}) = \sum x^{2}\,P(X = x)$, weighting the squared values by the same probabilities.
3. Subtract them, then take the square root for the standard deviation.

Note

The formula booklet gives $\operatorname{Var}(X) = E(X^{2}) - [E(X)]^{2}$ under Topic 4 (AHL).
The standard deviation is $\sigma = \sqrt{\operatorname{Var}(X)}$, back in the original units of $X$.

Example question

The discrete random variable $X$ has the probability distribution below.

x	1	2	3
P(X = x)	0.2	0.5	0.3

Find $E(X)$, $\operatorname{Var}(X)$ and the standard deviation of $X$.

Solution

$$E(X) = 1(0.2) + 2(0.5) + 3(0.3) = 2.1$$

$$E(X^{2}) = 1^{2}(0.2) + 2^{2}(0.5) + 3^{2}(0.3) = 0.2 + 2.0 + 2.7 = 4.9$$

$$\operatorname{Var}(X) = 4.9 - 2.1^{2} = 4.9 - 4.41 = 0.49$$

$$\sigma = \sqrt{0.49} = 0.7$$

From Sums To Integrals: The Probability Density Function

A continuous random variable can take any value in a range, like a waiting time or a length, so there are infinitely many possible values and you cannot list a probability for each one.
Instead a probability density function $f(x)$ describes how probability is spread out, and probability becomes area under the curve. $$P(a \le X \le b) = \int_{a}^{b} f(x)\,dx$$
1. $f(x) \ge 0$ everywhere, since area cannot be negative.
2. $\displaystyle\int_{-\infty}^{\infty} f(x)\,dx = 1$, since the total probability is 1.

Common Mistake

A density is not a probability, so $f(x)$ can be greater than 1; only the area under it is a probability.
The probability of any single exact value is an area of zero width, so $P(X = c) = 0$, and $P(X < c)$ equals $P(X \le c)$.
So strict and non-strict inequalities give the same probability for continuous variables.

Mean, Mode, Median And Variance For Continuous Variables

Every discrete formula carries over by replacing the sum $\sum$ with an integral $\int$ and the probability $P(X=x)$ with $f(x)\,dx$. $$E(X) = \int x\,f(x)\,dx, \qquad E(X^{2}) = \int x^{2} f(x)\,dx, \qquad \operatorname{Var}(X) = E(X^{2}) - [E(X)]^{2}$$
The mode: The value where $f(x)$ is greatest, the peak of the density, found by maximising $f$ and checking endpoints too.
The median $m$: The value that splits the area in half, so $\displaystyle\int_{-\infty}^{m} f(x)\,dx = \tfrac{1}{2}$.

Note

The formula booklet gives the continuous $E(X) = \int x f(x)\,dx$ and the variance, so those are provided.
The mode and median come from the density itself, found by maximising $f$ and by splitting the area, not from a booklet formula.

Example question

A continuous random variable $X$ has probability density function $f(x) = kx$ for $0 \le x \le 2$, and $f(x) = 0$ otherwise.

(a) Find $k$.

(b) Find the mean.

(d) State the mode.

(e) Find the variance.

Solution

(a): total area is 1, so $\int_{0}^{2} kx\,dx = k\left[\tfrac{x^{2}}{2}\right]_{0}^{2} = 2k = 1$.
$k = \tfrac{1}{2}$, so $f(x) = \tfrac{x}{2}$.
(b): $E(X) = \int_{0}^{2} x \cdot \tfrac{x}{2}\,dx = \left[\tfrac{x^{3}}{6}\right]_{0}^{2} = \tfrac{8}{6} = \tfrac{4}{3} \approx 1.33$.
(c): median $m$ solves $\int_{0}^{m} \tfrac{x}{2}\,dx = \tfrac{m^{2}}{4} = \tfrac{1}{2}$, so $m = \sqrt{2} \approx 1.41$.
(d): $f(x) = \tfrac{x}{2}$ increases across $[0, 2]$, so the mode is at the endpoint $x = 2$.
(e): $E(X^{2}) = \int_{0}^{2} x^{2}\cdot\tfrac{x}{2}\,dx = \left[\tfrac{x^{4}}{8}\right]_{0}^{2} = 2$, so $\operatorname{Var}(X) = 2 - \left(\tfrac{4}{3}\right)^{2} = \tfrac{2}{9} \approx 0.222$.

Linear Transformations: Shifting And Scaling A Variable

Changing units or adding a constant turns $X$ into $aX + b$, like converting Celsius to Fahrenheit or adding a fixed bonus to every score.
The mean and variance respond in a predictable way that works for both discrete and continuous variables. $$E(aX + b) = a\,E(X) + b \qquad\qquad \operatorname{Var}(aX + b) = a^{2}\operatorname{Var}(X)$$
1. Adding $b$ slides every outcome along by $b$, so the mean shifts by $b$ but the spread is unchanged.
2. Multiplying by $a$ scales the mean by $a$ and the variance by $a^{2}$, because variance is built from squared distances.
3. The standard deviation therefore scales by $|a|$, and the constant $b$ has no effect on it at all.

Exam technique

These two transformation results are not in the formula booklet, so learn them.
The two traps are forgetting to square the $a$ in the variance, and wrongly letting $b$ change the variance.

Example question

The variable $X$ from the earlier table has $E(X) = 2.1$ and $\operatorname{Var}(X) = 0.49$. Let $Y = 3X + 5$. Find $E(Y)$, $\operatorname{Var}(Y)$ and the standard deviation of $Y$.

Solution

$$E(Y) = 3E(X) + 5 = 3(2.1) + 5 = 11.3$$ $$\operatorname{Var}(Y) = 3^{2}\operatorname{Var}(X) = 9(0.49) = 4.41$$

$$\sigma_{Y} = \sqrt{4.41} = 2.1$$ which is 3 times $\sigma_{X}$, confirming the standard deviation scales by $|a|$.

Active recall

Write down the shortcut formula for the variance of a random variable in terms of $E(X^{2})$ and $E(X)$.
For a continuous random variable, what do you integrate to find a probability, and why is $P(X = c) = 0$?
How do you locate the median $m$ of a continuous random variable from its density $f(x)$?
State how $E(X)$ and $\operatorname{Var}(X)$ change under the transformation $aX + b$, and explain why $b$ does not affect the variance.
Give the two conditions a function must satisfy to be a valid probability density function.

IB Mathematics Analysis and Approaches AHL 4.14—properties of Discrete and Continuous Random Variables Notes

IB Mathematics Analysis and Approaches AHL 4.14—properties of Discrete and Continuous Random Variables Notes

Discrete And Continuous Random Variables: Spread, Densities And Transformations

Variance Of A Discrete Random Variable

From Sums To Integrals: The Probability Density Function

Mean, Mode, Median And Variance For Continuous Variables

Linear Transformations: Shifting And Scaling A Variable

Discrete And Continuous Random Variables: Spread, Densities And Transformations

Variance Of A Discrete Random Variable

From Sums To Integrals: The Probability Density Function

Mean, Mode, Median And Variance For Continuous Variables

Linear Transformations: Shifting And Scaling A Variable

Number and Algebra

Functions

Geometry & Trigonometry

Statistics & Probability

Calculus

IB Mathematics Analysis and Approaches AHL 4.14—properties of Discrete and Continuous Random Variables Notes

Number and Algebra

Functions

Geometry & Trigonometry

Statistics & Probability

Calculus

IB Mathematics Analysis and Approaches AHL 4.14—properties of Discrete and Continuous Random Variables Notes

Discrete And Continuous Random Variables: Spread, Densities And Transformations

Variance Of A Discrete Random Variable

From Sums To Integrals: The Probability Density Function

Mean, Mode, Median And Variance For Continuous Variables

Linear Transformations: Shifting And Scaling A Variable

Discrete And Continuous Random Variables: Spread, Densities And Transformations

Variance Of A Discrete Random Variable

From Sums To Integrals: The Probability Density Function

Mean, Mode, Median And Variance For Continuous Variables

Linear Transformations: Shifting And Scaling A Variable