Continuous Random Variables Revision Notes for Edexcel A-Level Further Mathematics

16.1.2 Continuous Random Variables

Continuous Random Variables

Discrete random variables can only take discrete values.
- Example: $X \sim B(3, 0.25), X \in \{0, 1, 2, 3\}$
- $P(X = 1.5)$ does not make sense.
Continuous random variables can take values across a continuous scale, allowing us to model real-life occurrences such as heights of trees, masses of babies, etc.

Functions of Continuous Random Variables

Given the probability density function of a continuous random variable $X$ , it is possible to find the probability density function or cumulative distribution function of $F(X)$ where $F$ is any function.

Worked Example

lightbulbExample

Example $X$ is a random variable that represents the length of the side of a square. The length of the side of the square is equally likely to take any value between 1 and 3.

Find the cumulative distribution function for the area of the square $A$ and hence its probability density function.

Step 1: Define the probability density function for the given variable:

f(x) = \begin{cases} \frac{1}{2} & \text{for } 1 \leq x \leq 3 \\ 0 & \text{otherwise} \end{cases}

Step 2: Determine the cumulative distribution function for this variable:

F(x) = \int_1^x \frac{1}{2} \, dx = \left[ \frac{1}{2}x \right]_1^x = \frac{1}{2}x - \frac{1}{2}

Thus:

F(x) = \begin{cases} 0, & x < 1 \\ \frac{1}{2}x - \frac{1}{2}, & 1 \leq x \leq 3 \\ 1, & x > 3 \end{cases}

Step 3: State the relationship between the two variables:

$X$ = length of the side of the square.
$A$ = area of the square. Thus:

A = X^2

Step 4: Define the cumulative distribution function ( $CDF$ ) of the target distribution in terms of probabilities:

$G(a) = P(A \leq a)$

Using $G$ to define the $CDF$ of $A$ .

Step 5: Rewrite the above cumulative distribution function in terms of the original variable. This rearranges to make the original variable the subject:

G(a) = P(X^2 \leq a) = P(-\sqrt{a} \leq X \leq \sqrt{a})

At this point, check whether the two limits make sense. In this case, $X$ can only take values between 1 and 3 from its initial definition.

= P(1 \leq X \leq \sqrt{a}) \quad \text{(This is G(A))}.

Step 6: Evaluate the probability using the original cumulative distribution function:

P(1 \leq X \leq \sqrt{a}) = F(\sqrt{a}) - F(1)

= \frac{{1}}{2}\sqrt a - \frac{1}{2} - 0 = \frac{1}{2} \sqrt{a} - \frac{1}{2}

(Remember to state the domain):

G(a) = \begin{cases} 0 & a < 1 \\ \frac{1}{2} \sqrt{a} - \frac{1}{2} & 1 \leq a \leq 9 \quad (\text{since } 1 \leq x \leq 3 \text{ and } A = X^2) \\ 1 & a > 9 \end{cases}

Step 7: If asked to find the probability density function, we find that $G'(A) = g(A)$ :

g(a) = G'(a) = \frac{d}{da} \left( \frac{1}{2} a^{\frac{1}{2}} - \frac{1}{2} \right) = \frac{1}{4} a^{-\frac{1}{2}}

Thus:

g(a) = \begin{cases} \frac{1}{4} a^{-\frac{1}{2}} & 1 \leq a \leq 9 \\ 0 & \text{otherwise} \end{cases}

infoNote

Past Paper Example

Q4 (June 2010, Q8)

The continuous random variable $S$ has a probability density function given by:

f(s) = \begin{cases} \frac{8}{3s^{-3}} & 1 \leq s \leq 2 \\ 0 & \text{otherwise} \end{cases}

An isosceles triangle has equal sides of length $S$ , and the angle between them is 30°. (See diagram.)

(i) Find the (cumulative) distribution function of the area $X$ of the triangle, and hence show that the probability density function of $X$ is $\frac{1}{3x^2}$ over an interval to be stated.

(ii) Find the median value of $X$ .

Solution: i)

f(s) = \frac{8}{3} s^{-3} \quad \Rightarrow \quad F(s) = \int_1^s \frac{8}{3} s^{-3} \, ds = \left[ -\frac{4}{3} s^{-2} \right]_1^s

= -\frac{4}{3} s^{-2} + \frac{4}{3} = \frac{4}{3} - \frac{4}{3} s^{-2}

Thus:

F(s) = \begin{cases} 0 & s < 1 \\ \frac{4}{3} - \frac{4}{3} s^{-2} & 1 \leq s \leq 2 \\ 1 & s > 2 \end{cases}

The area of the triangle is given by:

X = \frac{1}{2} S^2 \sin 30^\circ = \frac{1}{4} S^2

Thus:

G(x) = P(X \leq x) = P\left( \frac{1}{4} S^2 \leq x \right)

= P(S^2 \leq 4x) = P(-2\sqrt{x} \leq S \leq 2\sqrt{x})

= P(1 \leq S \leq 2\sqrt{x}) \quad \text{since} \quad 1 \leq S \leq 2

\Rightarrow G(x) = F(2\sqrt{x}) - \cancel{F(1)}

= \frac{4}{3} - \frac{4}{3}(2\sqrt{x})^{-2} = \frac{4}{3} - \frac{\cancel{4}}{3} \times \frac{1}{\cancel 4x} = \frac{4}{3} - \frac{1}{3x}

Thus:

G(x) = \begin{cases} 0 & x < \frac{1}{4} \\ \frac{4}{3} - \frac{1}{3x} & \frac{1}{4} \leq x \leq 1 \\ 1 & x > 1 \end{cases}

From the given diagram, we note that $1 \leq S \leq 2$ and $x = \frac{1}{4} S^2$ .

Thus:

G(x) = \frac{4}{3} - \frac{1}{3}x^{-1} \Rightarrow G'(x) = \frac{1}{3}x^{-2}=\frac {1}{3x^2} \quad (\text{as required})

Thus:

g(x) = \begin{cases} \frac {1}{3x^2} & \frac{1}{4} \leq x \leq 1 \\ 0 & \text{otherwise} \end{cases}

Solution: (ii) Find the median value of $x$ .

The median is $x$ such that $G(x) = 0.5$ :

\frac{4}{3} - \frac{1}{3x} = 0.5 \quad \Rightarrow \quad -\frac{1}{3x} = -\frac{5}{6} \quad \Rightarrow \quad 1 = \frac{15x}{6}

Thus:

x = \frac{6}{15} = \frac{2}{5} \quad \Rightarrow \text{Median is} \quad \frac{2}{5}

Mean and Variance of Continuous Random Variables (CRVs)

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Recap of Mean and Variance of Discrete Random Variables:

$\text{Mean} = E(X) = \sum x p$

$\text{Variance} = E(X^2) - E(X)^2 = \sum x^2 p - \left( \sum x p \right)^2$

For CRVs, the concept of mean and variance does not change.

To calculate E(X) , we take all the x values multiplied by the corresponding p -values, then calculate the sum.

As the number of strips becomes infinite, $\sum x p(x)$ becomes $\int x p(x) dx .$

For CRVs:

$E(X) = \int x p(x) dx$

$E(X^2) = \int x^2 p(x) dx$

$\text{Var}(X) = E(X^2) - E(X)^2$

Where p(x) is the probability density function.

infoNote

Worked Example

a) $E(Y) = \int_0^4 y f(y) dy = \int_0^4 y \frac{y}{8} dy = \frac{1}{8} \int_0^4 y^2 dy = \frac{[y^3/3]}{8} \Bigg|_0^4$

$= \frac{4^3}{24} = \frac{64}{24} = \frac{8}{3}$

b) $\text{Var}(Y) = E(Y^2) - E(Y)^2$

$E(Y^2) = \int_0^4 y^2 f(y) dy = \frac{1}{8} \int_0^4 y^3 dy = \frac{[y^4/4]}{8} \Bigg|_0^4 = \frac{4^4}{32} = \frac{256}{32} = 8$

$\text{Var}(Y) = 8 - \left(\frac{8}{3}\right)^2 = 8 - \frac{64}{9} = \frac{72}{9} - \frac{64}{9} = \frac{8}{9}$

c) Remember: $\sigma = \sqrt{\text{Variance}}$

$\sigma = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}$

d) $P(Y > \mu)$

$P(Y > \frac{8}{3}) = \int_{8/3}^4 \frac{y}{8} dy = \frac{1}{16} \Bigg[ y^2 \Bigg|_{8/3}^4 \Bigg] = \frac{4^2}{16} - \left(\frac{8}{3}\right)^2$

$= \frac{16}{16} - \frac{64}{9 \times 16} = \frac{5}{9}$

e) $\text{Var}(3Y + 2) = 3^2 \times \text{Var}(Y) = 9 \times \frac{8}{9} = 8$

f) $E(Y + 2) = \frac{8}{3} + 2 = \frac{14}{3}$

Remember:

$\text{Var}(aXe + b) = a^2 \text{Var}(X), \quad E(aXe + b) = aE(X) + b$

Median, Mode, and Skewness of CRV

Mode

The mode of a continuous random variable (CRV) is the maximum of its probability density function (if a maximum exists).

lightbulbExample

Example The continuous random variable $X$ has a probability density function given by:

f(x) = \begin{cases} \frac{3}{80} (8 + 2x - x^2) & \text{for } 0 \leq x \leq 4, \\ 0 & \text{otherwise.} \end{cases}

Tasks

a) Sketch the probability density function of $X$ .

b) Find the mode of $X$ .

a) $f(x) = \frac{3}{80}(x^2 - 2x - 8) = \frac{-3}{80}(x - 4)(x + 2)$

                                      (Root at $ x = 4$  or  $x = -2$

$f(0) = \frac{-3}{80}(8) = \frac{3}{10}$

LHS (left-hand side) is in $(0, \frac{3}{10})$ .

$f(4) = \frac{-3}{80}(4^2 - 2(4) - 8) = 0$

RHS (right-hand side) is $(4, 0)$ .

b) The mode may or may not be a local maximum as found by differentiation.

In this P.D.F., differentiating and finding the stationary point would lead to a mode outside of the domain of validity of the P.D.F. Here, from observing where the function is valid, we see the mode is f(0) .

(In this question, the mode is found by differentiation.)

$f(x) = \frac{3}{80} (8 + 2x - x^2) \quad 0 \leq x \leq 4 \quad \text{otherwise}$

(i.e., mode on this function is the stationary point)

$f'(x) = \frac{3}{80} (2 - 2x) = 0$

$\Rightarrow 2 = 2x \Rightarrow x = 1$

Mode = $x -value$ point with the highest frequency.

:::

Median

The median is the value of $x$ to the left of which 0.5 of the probability lies and to the right of which 0.5 of the probability lies.

Method 1: Using P.D.F.

Method 2: Use C.D.F.

Solve $F(x) = 0.5$

lightbulbExample

Example: The continuous random variable $X$ has a cumulative distribution function given by:

F(x) = \begin{cases} 0 & x < 0, \\ \frac{x^2}{6} & 0 \leq x \leq 2, \\ -\frac{x^2}{3} + 2x - 2 & 2 \leq x \leq 3, \\ 1 & x > 3. \end{cases}

Tasks

a) Find the median value of $X$ .

b) Find the quartiles and the interquartile range (IQR) of $X$ .

Note: Ensure answers are given to 3 decimal places.

a) $F(x) = 0.5$

$0.5 = \frac{x^2}{6} \Rightarrow x^2 = 3 \Rightarrow x = \sqrt{3}$

(Since $\sqrt{3} \approx 1.732$ and $0 \leq x \leq 2$ , this is valid.)

b) $F(x) = 0.75$ (Upper quartile)

$0.75 = \frac{x^2}{6} \Rightarrow x^2 = 4.5 \Rightarrow x = \frac{3\sqrt{2}}{2}$

(But this is not valid, so we use the second part of the function.)

$F(x) = \frac{-x^3}{3} + 2x - 2 = 0.75$

$\Rightarrow -\frac{x^3}{3} + 2x - 2.75 = 0 \Rightarrow x = \frac{6 - \sqrt{13}}{2} \approx 2.154$

Thus, $\text{UQ} = \frac{6 - \sqrt{13}}{2} \approx 2.154 .$

$LQ:$ $F(x) = 0.25$

$0.25 = \frac{x^2}{6} \Rightarrow x^2 = \frac{3}{2} \Rightarrow x = \frac{\sqrt{6}}{2} \approx 1.225$

Thus, $\text{LQ} = \frac{\sqrt{6}}{2} \approx 1.225$

Interquartile Range (IQR):

$IQR = \frac{6 - \sqrt{3}}{2} - \frac{\sqrt{6}}{2} \approx 0.909$

Continuous Random Variables (Edexcel A-Level Further Mathematics): Revision Notes

16.1.2 Continuous Random Variables