# Expected Value and Variance of Exponential Random Variable

## Problem 757

Let $X$ be an exponential random variable with parameter $\lambda$.

**(a)** For any positive integer $n$, prove that

\[E[X^n] = \frac{n}{\lambda} E[X^{n-1}].\]

**(b)** Find the expected value of $X$.

**(c)** Find the variance of $X$.

**(d)** Find the standard deviation of $X$.

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## Solution.

### Solution of (a)

Recall that the probability density function $f(x)$ of an exponential random variable with parameter $\lambda$ is given by

\begin{align*}

f(x) =

\begin{cases}

\lambda e^{-\lambda x} & \text{ if } x \geq 0\\

0 & \text{ if } x < 0
\end{cases}
\end{align*}
and the parameter $\lambda$ must be positive.
It follows from this and the definition of expectation, we get
\begin{align*}
E[X^n] &= \int_0^{\infty} x^n \cdot \lambda e^{-\lambda x} dx.
\end{align*}
Applying integral by parts with
\[u = x^n, dv=\lambda e^{-\lambda x} dx\]
and hence
\[du = nx^{n-1}dx, v = -e^{-\lambda x},\]
we obtain (from $\int u dv = uv - \int v du$)
\begin{align*}
E[X^n] &= \int_0^{\infty} x^n \cdot \lambda e^{-\lambda x} dx\\[6pt]
&= \left[x^n \cdot (-e^{-\lambda x})\right]_0^{\infty} - \int_0^{\infty} (-e^{\lambda x}) \cdot nx^{n-1} dx\\[6pt]
&= 0 + n \int_0^{\infty} e^{\lambda x} x^{n-1} dx\\[6pt]
&= \frac{n}{\lambda} \int_0^{\infty} x^{n-1} \cdot \lambda e^{\lambda x} dx\\[6pt]
&= \frac{n}{\lambda} E[X^{n-1}].
\end{align*}
This proves the required equality
\[E[X^n] = \frac{n}{\lambda} E[X^{n-1}].\]

### Solution of (b)

The expected value $E[X]$ can be obtained from the formula we just proved in part (a) by substituting $n=1$. Thus, we have

\begin{align*}

E[X] = \frac{1}{\lambda} E[1] = \frac{1}{\lambda}.

\end{align*}

### Solution of (c)

We calculate the variance using the formula

\[V(X) = E[X^2] – (E[X])^2.\]
We know $E[X] = 1/\lambda$ from part (b). To compute $E[X^2]$, let $n=2$ in the formula in part (a). Then

\begin{align*}

E[X^2] &= \frac{2}{\lambda}E[X]\\[6pt]
&= \frac{2}{\lambda} \cdot \frac{1}{\lambda}\\[6pt]
&= \frac{2}{\lambda^2}.

\end{align*}

Combining these, we obtain

\begin{align*}

V(X) &= E[X^2] – (E[X])^2\\[6pt]
&= \frac{2}{\lambda^2} – \frac{1}{\lambda^2}\\[6pt]
& = \frac{1}{\lambda^2}.

\end{align*}

Therefore, the variance of $X$ is

\[V(X) = \frac{1}{\lambda^2}.\]

### Solution of (d)

Taking the square root of the variance $V(X)$, we obtain the standard deviation

\[\sigma = \frac{1}{\lambda}.\]

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