Limits of integration of manipulated probability distribution function

Question

I have to compute $\mathbb E\left[e^X \right]$, where $X$ follows a uniform distribution on $[0,1]$.

I have started by computing the probability density function of $Y = e^X$, by doing the following: $$ F_Y(x) = \mathbb P(Y \leq x) = \mathbb P(e^X\leq x) = \mathbb P(X\leq \log(x)) = \int_0^{\log(x)}f(y)\,dy = \log(x) $$ where $f(x)$ is the probability density function of the uniform random variable.

Now, in the final step, when I have to compute the expectation $\mathbb E[Y]$, I am unsure which limits of integration to choose.

I have the feeling it is no longer $0$ and $1$, and that is the question precisely. How do the limits of integration change, when changing the probability distribution function? (for example, in this case, the random variable has exponentiated).

Answer 1

This question is easier to do directly so I'll do both (btw. this is just the moment generating function of the random variable with $t=1$ http://en.wikipedia.org/wiki/Moment-generating_function).

$X$ can take values in $[0,1]$ so $e^X$ can take values in $[e^0,e^1]=[1,e]$. Let's now use your work and see what we get, \begin{align*} F_Y(x)=log(x) \Rightarrow f_Y(x)=\frac{1}{x}\\ E[Y]=\int_1^e x\frac{1}{x}dx=x|_1^e=e-1 \end{align*} Doing the problem directly now (as @V.C. just did), \begin{align*} E[Y]=E[e^X]=\int_0^1e^xdx=e^x|^1_0=e-1 \end{align*}

Answer 2

We can use the law of the unconscious statistician to calculate $\mathbb E[e^X]$. We have that $$ \mathbb E[e^X]=\int_{-\infty}^{\infty}e^xf(x)\mathrm dx=\int_0^1e^x\mathrm dx=e-1 $$ since the density function $f$ of the uniform distribution is equal to $1$ if $0\le x\le1$ and equal to $0$ otherwise.