Solution to Exercise 1.

(1) [20 points] Suppose $X$ and $Y$ are independent random variables with the distributions as given below. For each case, determine the distribution of $Z$ $=$ $XY$ either by giving the cdf, pdf, or pmf. (a) $X$ is uniform(0,1) and $Y$ is exponential ($\mu$).

Solution: Since the joint distribution of $(X,Y)$ is continuous, we work with the pdf

$$f_{XY} (x,y) \; = \; I_{(0,1)} (x) \mu^{-1} \exp[-y/\mu] I_{(0,\infty)} (y) ,$$

and apply the ``Jacobian'' theory. The transformation we will use is

$$(W,Z) \; = \; (Y,XY) \quad \Longrightarrow \quad (X,Y) \; = \; (Z/W,W) .$$

The Jacobian is

$$\left\vert \frac{\partial (x,y)}{\partial (w,z)} \right\ver... ... (-z/w^2) \cdot 0 - (1/w) \cdot 1 \right\vert \; = \; 1/w .$$

Substituting in the joint pdf, we have

$$\begin{eqnarray*} f_{WZ} (w,z) & \; = \; & f_{XY} (z/w,w) \cdot (1/w) \\ & \;... ...{(0,1)} (z/w) \mu^{-1} w^{-1} \exp[-w/\mu] I_{(0,\infty)} (w) . \end{eqnarray*}$$

Note that $I_{(0,1)} (z/w) I_{(0,\infty)} (w)$ is $1$ if and only if $0 < z/w < 1$ $\Longleftrightarrow$ $0 < z < w < \infty$. For our purposes, we note that

$$I_{(0,1)} (z/w) I_{(0,\infty)} (w) \; = \; I_{(z,\infty)} (w) I_{(0,\infty)} (z) .$$

Now we want the marginal pdf $f_Z (z)$, which is

$$f_Z (z) \; = \; \int \, f_{WZ} (w,z) \, dw$$

$$\; = \; \int_z^{\infty} \, \mu^{-1} w^{-1} \exp[-w/\mu] \, dw I_{(0,\infty)} (z)$$ (1)

$$\mbox{ change variables } v = w/z$$

$$\; = \; \mu^{-1} \int_{1}^{\infty} \, v^{-1} \exp[-vz/\mu] \, dw I_{(0,\infty)} (z)$$

$$\; = \; \mu^{-1} \mbox{Ei}(1,z/\mu) I_{(0,\infty)} (z) ,$$

where $\mbox{Ei}(n,y)$ $=$ $\int_1^{\infty} x^{-n} \exp[-yx] dx$ is a well known special function called the Exponential Integral. (What? You don't know this special function? No problem. I didn't count off as long as you got a correct integral expression for $f_Z (z)$ as in (1) or (2) below.) An alternative expression (if you chose the transformation $(W,Z) \; = \; (X,XY)$ is

$$f_Z (z) \; = \; \int_0^1 \, (\mu w)^{-1} \exp \left[ -z / (w \mu) \right] \, dw I_{(0,\infty)} (z) .$$ (2)

(b) $X$ is uniform(0,1) and $Y$ is Bernoulli(1/2).

Solution: Here, $X$ has a continuous distribution but $Y$ is discrete. We can't use a Jacobian argument here. Also, we expect that the distribution of $Z = XY$ might be a mixture of discrete and continuous, so we won't have a pdf or a pmf. It will be necessary to give the distribution using the cdf. Clearly, $Y=0$, which occurs with probability $1/2$, means $Z = 0$. If $Y = 1$, then $Z = X$. Thus, $P[ Z = 0] = 1/2$, and if $0 < z < 1$, then

$$F_Z (z) \; = \; P[ Z \le z ]$$

$$\; = \; P[ Z \le z \vert Y = 0 ] P[ Y = 0 ] \, + \, P[ Z \le z \vert Y = 1 ] P[ Y = 1 ]$$

$$\; = \; 1 \cdot (1/2) \, + \, P[ X \le z ] \cdot (1/2)$$

$$\; = \; (1/2)(1+z) .$$ (3)

Putting it together, we have

$$F_z (z) \; = \; \left\{ \begin{array}{ccl} 0 & & \mbox{i... ... \le z \le 1\\ 1 & & \mbox{if } z > 1. \end{array} \right.$$ (4)

(c) $X$ is Poisson(1) and $Y$ is Bernoulli(1/2).

Solution: Both $X$ and $Y$ are discrete, so we don't use any Jacobians. The distribution of $Z$ will be discrete with possible values $\{ 0, 1 , \ldots \}$, the nonnegative integers. The pmf is

$$\begin{eqnarray*} f_Z (z) & \; = \; & P[ Z = z ] \\ & \; = \; & P[Z=z\vert Y... ...& & \mbox{if } z \in \{ 1 , 2 , \ldots \}. \end{array} \right. \end{eqnarray*}$$