1. 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.

    2. 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 \(1/\lambda\).

  2. For which value of \(x\) the density of the exponential distribution reaches its maximum value? What is this value?

    3. 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 \(\lambda \exp(-0) = \lambda\).