I am trying to find the expected value of $Z$ where $Z = Y\cdot e^X$ where $Y \sim U(0,1)$ and $X \sim U(0,1)$.

My attempt so far:

$$F_Z(z) = P(Ye^X \le z) = \int \int_{Ye^X \le z} f(x,y)\, dxdy$$

Where $f_{X,Y}(x,y) = f_y\cdot f_{e^x}$

$$f_Y(y) = \frac{1}{1-0}\,, \quad y \in (0,1)$$

I am stuck trying to find $f_{e^X}$ but I cannot remember how to find that pdf.

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    $\begingroup$ Hint: Focus on expectations only and add the self-sfudy tag. $\endgroup$ – Michael M Oct 25 '13 at 19:12
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    $\begingroup$ If $x$ is independent of $Y$, then (roughly speaking) any function of $x$ is also independent of $Y$. In particular $f(x)=e^x$ is independent of $Y$. $\endgroup$ – Stat Oct 25 '13 at 19:12
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    $\begingroup$ It is a good idea to get in the habit of distinguishing random variables from their possible values or realizations: it will help avoid misunderstandings when problems get more complicated than this. You could begin by editing this question to use the variables $x$, $y,$ $z,$ $X,$ $Y,$ and $Z$ in a logical and consistent way. (The convention is to use capital letters for the random variables.) $\endgroup$ – whuber Oct 25 '13 at 19:48
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    $\begingroup$ Hint: when $(X,Y)$ have a joint pdf $f$, isn't the definition of the expectation of a bivariate function $g$, such as $g(X,Y)=Y\exp(X),$ given by $\iint g(x,y)f(x,y)dxdy$? $\endgroup$ – whuber Oct 25 '13 at 19:52

If the problem does not state explicitly that $X$ and $Y$ are independent, then it doesn't have a solution, because the marginal distributions of $X$ and $Y$ do not determine their joint distribution.

Supposing that $X$ and $Y$ are independent, then $e^X$ and $Y$ are also independent. Proof:

$$ \begin{eqnarray} P\left\{e^X \in A, Y\in B \right\} &=& P\left\{X \in \exp^{-1}(A), Y\in B \right\} \\ &=& P\left\{X \in \exp^{-1}(A)\right\}P\left\{Y\in B \right\} \quad \textrm{[independence of $X$ and $Y$]} \\ &=& P\left\{e^X \in A\right\}P\left\{Y\in B \right\} \, , \end{eqnarray} $$ in which $A$ and $B$ are Borel sets, and $\exp^{-1}(A)$ is the inverse image of the set $A$ under $\exp$. Note that $\exp^{-1}(A)$ is also a Borel set because $\exp:\mathbb{R}\to\mathbb{R}$ is measurable.

But the expectation of a product of independent random variables is the product of their expectations. Therefore, $$ \mathrm{E}\left[Ye^X\right]=\mathrm{E}[Y]\cdot\mathrm{E}\left[e^X\right] \, . $$ To compute $\mathrm{E}\left[e^X\right]$ you don't need to determine the distribution of $Z=e^X$. It follows from a theorem, known folkloricaly as the Law of the Unconscious Statistician, that $$ \mathrm{E}\left[e^X\right] = \int_{-\infty}^\infty e^x f_X(x)dx \, , $$ and this integral is easy to compute.


There is definitely an easier approach to this problem (hints were given in the comments), but since you asked about a specific step, I'll go from there.

You want to compute the pdf $f_{e^X}(x)$. Let's start with:

$$ F_{e^X}(x) = P(e^X < x) = P(X < \log x) = (F_X \circ \log)(x) $$

Recall that we can compute $f_{e^X}$ by differentiating $F_{e^X}$. In this case, you can use the chain rule.

This kind of transformation generalizes to the multivariate case.


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