# Chapter 4 - Continuous Random Variables and Probability DistributionsEdit

## Section 4.1Edit

### Exercise 1Edit

Given the density function $f(x) = \begin{cases} 0.5x & 0 \leq x \leq 2 \\ 0 & \text{otherwise} \end{cases}$

• Part a. Find $P(X \leq 1)$

$\begin{array}{rl} P(X \leq 1) &= \int_{-\infty}^1 f(x) dx \\ &= \int_{0}^1 0.5 x dx = \left. \frac{x^2}{4} \right|_0^1 \\ &= \frac{1}{4} \end{array}$

• Part b.

$\begin{array}{rl} P(0.5 \leq X \leq 1.5) &= \int_{0.5}^{1.5} f(x) dx = \left. \frac{x^2}{4} \right|_{0.5}^{1.5} \\ &= \frac{1}{2} \end{array}$

• Part c.

$\begin{array}{rl} P(1.5 < X ) &= \int_{0.5}^{2} f(x) dx = \left. \frac{x^2}{4} \right|_{0.5}^{2} \\ &= 0.438 \end{array}$

### Exercise 2Edit

Let $X \text{ be Uniform}(-5,5)$

• Part a.

$\begin{array}{rl} P(X < 0) &= \int_{-5}^0 \frac{1}{5- -5} dx \\ \\ &= \left. \frac{x}{10} \right|_{-5}{0} \\ \\ &= \frac{1}{2} \end{array}$

• Part b.

$P(-2.5 < X < 2.5) = \int_{-2.5}^{2.5} \frac{1}{10} dx = \frac{1}{2}$

• Part c.

$P(-2 < X < 3) = \int_{-2}^{3} \frac{1}{10} dx = \frac{1}{2}$

• Part d. For $-5 < k < k + 4 < 5$, compute

$\begin{array}{rl} P(k < X < k+4) &= \int_{k}^{k+4} \frac{1}{10} dx \\ \\ &= \left. \frac{x}{10} \right|_{k}^{k+4} \\ \\ &= \frac{k+4}{10} - \frac{k}{10} \\ \\ &= \frac{4}{10} = \frac{2}{5} \end{array}$

### Exercise 3.Edit

Let $f(x)$ be a probability density function.

$f(x) = \begin{cases} 0.09375(4-x^2) & -2 \leq x \leq 2 \\ 0 & \text{otherwise} \end{cases}$
• Part a. graph $f(x)$

• Part b.

$P(X > 0) = \int_{0}^{2} f(x) dx = 0.5$

• Part c.
$P(-1 < X < 1) = \int_{-1}^{1} f(x) dx = 0.6875$
• Part d.
$P(X < -0.5 \text{ or } X > 0.5) = 1 - P(-5 \leq X \leq 0.5) = 1 - \int_{-0.5}^{0.5} f(x) dx = 1 - 0.3671888 = 0.632812$

### Exercise 4.Edit

Let $X$ have the Rayleigh distribution with the probability density function

$f(x) = \begin{cases} \frac{x}{\theta^2} e^{-x^2/\left(2\theta^2\right)} & x > 0 \\ & \\ 0 & \text{otherwise} \end{cases}$

• Part a.

Verify that $f(x)$ is a pdf.

• First notice that $f(x) \geq 0$ for all $x$
• Next show the integral over the whole number line equals one:
$\begin{array}{rll} \int_{-\infty}^{\infty}f(x)dx &= \int_{0}^{\infty} \frac{x}{\theta^2} e^{-x^2/\left(2\theta^2\right)} dx & \text{let } y = x/\theta, dy = 1/\theta dx \\ \\ &= \int_{0}^{\infty} y e^{-y^2/2} dy & \\ \\ &= \left. e^(-y^2/2) \right|_{0}^{\infty} \\ \\ &= 0 - - 1 \\ &= 1 \end{array}$
• Part b. Let $\theta = 100$.
• Probability $X$ is at most 200
$P(X \leq 200) = \int_{0}^{200} f(x) dx = 1 - \frac{1}{e^2} \approx 0.864665$
• Probability $X$ is less than 200
$P(X < 200) = \int_{0}^{200} f(x) dx = 1 - \frac{1}{e^2} \approx 0.864665$
• Probability $X$ is at least 200
$P(X \geq 200) = 1 - P(X < 200) = \frac{1}{e^2} \approx 0.135335$

• The probability $X$ is between 100 and 200 assuming $\theta = 100$.
$P(100 < X < 200) = \int_{100}^{200} f(x)dx = \frac{e^{3/2} - 1}{e^2} \approx 0.471195$
• Give an expression for $P(X \leq x)$, i.e., define the cumulative distribution function.
$P(X \leq x) = \int_{-\infty}^{x} f(y)dy = \begin{cases} 0 & x \leq 0 \\ 1 - \exp\left( \frac{-x^2}{2\theta^2} \right) & x > 0 \end{cases}$