12.4: Exponential and normal random variables
Exponential density function

Given a positive constant $k > 0$, the exponential density function (with parameter $k$) is

$$f(x) = \begin{cases}k e^{-kx} \text{ if } x \geq 0 \\ 0 \text{ if } x < 0 \end{cases}$$

Expected value of an exponential random variable

Let $X$ be a continuous random variable with an exponential density function with parameter $k$.

Integrating by parts with $u = kx$ and $dv = e^{-kx} dx$ so that $du = k dx$ and $v = \frac{-1}{k} e^{-kx}$, we find

$$\begin{eqnarray*} E(X) & = & \int_{-\infty}^\infty x f(x) dx \\ & = & \int_{0}... ...e^{-kx} - \frac{1}{k} e^{-kx}]\vert_{0}^r \\ & = & \frac{1}{k} \end{eqnarray*}$$

Variance of exponential random variables

Integrating by parts with $u = k x^2$ and $dv = e^{-kx} dx$ so that $du = 2k x dx$ and $v = \frac{-1}{k} e^{-kx}$, we have

$$\begin{eqnarray*} \int_0^\infty x^2 e^{-kx} dx & = & \lim_{r \to \infty}( [- x^2... ...-kx} - \frac{2}{k^2} e^{-kx}]\vert_0^r) \\ & = & \frac{2}{k^2} \end{eqnarray*}$$

So, $\mathrm{Var}(X) = \frac{2}{k^2} - E(X)^2 = \frac{2}{k^2} - \frac{1}{k^2} = \frac{1}{k^2}$.

Example

Exponential random variables (sometimes) give good models for the time to failure of mechanical devices. For example, we might measure the number of miles traveled by a given car before its transmission ceases to function. Suppose that this distribution is governed by the exponential distribution with mean $100,000$. What is the probability that a car's transmission will fail during its first $50,000$ miles of operation?

Solution

$$\begin{eqnarray*} \mathrm{Pr}(X \leq 50,000) & = & \int_0^{50,000} \frac{1}{100,... ...00} \\ & = & 1 - e^{\frac{-1}{2}} \\ & \approx & .3934693403 \end{eqnarray*}$$

Normal distributions

The normal density function with mean $\mu$ and standard deviation $\sigma$ is

$$f(x) = \frac{1}{\sigma \sqrt{2 \pi}} e^{\frac{-1}{2} (\frac{x - \mu}{\sigma})^2}$$

As suggested, if $X$ has this density, then $E(X) = \mu$ and $\mathrm{Var}(X) = \sigma^2$.

The standard normal density function is the normal density function with $\mu = \sigma = 1$. That is,

$$g(x) = \frac{1}{\sqrt{2 \pi}} e^{\frac{-1}{2} x^2}$$

Taylor expansion for the normal cumulative distribution function

Let $f(x) = \frac{1}{\sqrt{2 \pi}} e^{\frac{-1}{2} x^2}$ be the standard normal density function and let $F(x) = \int_{-\infty}^x f(t) dt$ be the standard normal cumulative distribution function.

We compute a Taylor series expansion,

$$\begin{eqnarray*} G(x) & = & \int \frac{1}{\sqrt{2 \pi}} e^{\frac{-1}{2} x^2} dx... ...c{1}{6} x^3 + \frac{1}{40} x^5 - \frac{1}{336} x^7 + \cdots ) \end{eqnarray*}$$

So $F(x) = G(x) + C$ for some $C$. As $0$ is the expected value, we need $\frac{1}{2} = F(0) = G(0) + C = C$.

Example

If the continuous random variable $X$ is normally distributed, what is the probability that it takes on a value of more than a standard deviations above the mean?

Solution

Via a change of variables, we may suppose that $X$ is normally distributed with respect to the standard normal distribution. Let $F$ be the cumulative distribution function for the standard normal distribution.

$$\begin{eqnarray*} \mathrm{Pr}(X \geq 1) & = & \int_1^\infty \frac{1}{\sqrt{2 \pi... ...\frac{1^7}{336}) \\ & \approx & 0.5 - .34332 \\ & = & .15668 \end{eqnarray*}$$