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The expected value ${\Bbb E}_{{\mathrm x} \sim p}[x]$ of a continuous distribution with probability density function given by $p$ is the "average" value of a random variable with this distribution, and can be computed using the following formula:

${\Bbb E}_{{\mathrm x} \sim p}[x] = \int_{-\infty}^{\infty} x p(x) dx$ .

Compute the expectation of the exponential distribution, whose density function is:

$p(x; \lambda) = \lambda (1_{x \geq 0}) \exp(-\lambda x)$ ,

for a given parameter $\lambda > 0$ , where the indicator function $1_{x \geq 0}$ is 1 if $x$ is non-negative and 0 if $x$ is negative.

A.:

We have:

$\begin{eqnarray*} {\Bbb E}_{{\mathrm x} \sim p}[x] &=& \int_{-\infty}^{\infty}x\lambda(1_{x \geq 0})\exp(-\lambda x)dx\\ &=& \int_{0}^{\infty} x \lambda \exp(-\lambda x) dx. \end{eqnarray*}$

Using integration by parts,

$\begin{eqnarray*} \left.\int_{0}^{\infty} x \lambda \exp(-\lambda x) dx = -x\exp(-\lambda x)\right|_0^{\infty} &-& \int_{0}^{\infty} \lambda \exp(-\lambda x) dx. \end{eqnarray*}$

The first term vanishes, because both extremes evaluate to zero. For the second term, we have:

$\begin{eqnarray*} -\int_{0}^{\infty} \lambda \exp(-\lambda x) dx &=& \left.-\frac{1}{\lambda}\exp(-\lambda x)\right|_0^{\infty} = \frac{1}{\lambda}. \end{eqnarray*}$

The expected value is therefore $1/\lambda$ .

For which value of $x$ the density of the exponential distribution reaches its maximum value? What is this value?

A.:

The density function is positive for $x \geq 0$ only. In this domain, the density is strictly decreasing. It follows that the maximum value is found for $x = 0$ , and the value is $\lambda \exp(-0) = \lambda$ .