Suppose $X$ is uniformly distributed on $(0, 1)$. Compute $E(e^X)$. Find the probability density function for $Y = e^X$ and use this to verify your answer.
I'm a little confused by how to solve this problem. I know if $X$ is uniformly distributed, then $f(x) = \frac{1}{b-a}$ or $1$ on the interval $(0,1)$ and 0 otherwise. I know the formula for the expected value of this is $\int_{a}^{b}xf(x) dx = \int_{0}^{1}xf(x) xdx$. I thought because in this case our $f(x)$ seems to be $e^x$, the expected value of $x$ would be $\int_{0}^{1}e^xf(x) = e-1$.
However, when I find the probability density function for $Y = e^x$, I go through the process of finding the cumulative density function and then differentiating and find that the probability density function $f_Y(y) = \frac{1}{y}$ for $1 < y < e$ and 0 elsewhere.
It seems like these two answers should be lining up, but they aren't. Am I missing something or was my thinking wrong for any part of this problem?
Lets take a look:
$$f_Y(y) = \frac{1}{y}$$
$$E(Y)=\int_{1}^{e}yf(y) dy=\int_{1}^{e}y \frac 1y dy=\int_{1}^{e}dy=e-1$$