Expected Value and Variance of Exponential Random Variable

Probability problems

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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