[Math] expected value of an exponential distribution
exponential distributionprobabilitystatistics
I've been stuck at this problem for hours now. So first I know that the expected value of $X\sim\operatorname{Exponential}(\lambda)$ is $1/\lambda.$ But I can't figure out what the expected value is given this information:
$X\sim\operatorname{Exponential}(3)$ and $Y=\exp(2x)$
So, how do I calculate the $\operatorname E[Y]$? A step by step would be helpful, I've been stuck for hours.
Best Answer
First, if $X \sim Exp(\lambda)$, then it has a p.d.f. given by $f_\lambda(x) = \lambda \mathrm{e}^{-\lambda x}$. By the definition of expectation:
$$ \mathrm{E}[Y] = \mathrm{E}[\mathrm{e}^{2X}] = \int_0^\infty \mathrm e^{2x} f_3(x) \mathrm{d} x = 3 \int_0 ^\infty \mathrm{e}^{2x}\mathrm{e}^{ - 3 x}\mathrm d x = 3 \int_0^\infty \mathrm{e}^{-x}\mathrm{d}x = 3.$$
Start with the joint distribution of $X_1$ and $X_2$, which will be a distribution over the (cartesian) product of the ranges of $X_1$ and $X_2$, i.e. over $(0,\infty)^2$. The area that you're interested in are those points $(x_1,x_2) \in (0,\infty)^2$ with $x_2 > x_1$.
$E[X_2|X_2 > X_1]$ is thus $$\begin{eqnarray}
\int_0^\infty \int_{x_1}^\infty x_2f_1(x_1)f_2(x_2) dx_2x_1
\end{eqnarray}
$$ where $f_1$ and $f_2$ are the densities of your two exponential distributions.
An exponential distribution of a random variable $X$ of expected value $1$ (i.e., mean) has PDF $f_X(x) = e^{-x}$ for $x \gt 0$, and zero otherwise. You want $E(\sqrt{X})$.
In general, to compute $E[(g(X)]$ for a distribution $f_X(x)$ is
$$\int_{-\infty}^{\infty} dx \, g(x) f(x) $$
In your case, the integral is
$$\int_0^{\infty} dx \, \sqrt{x} \, e^{-x}$$
To evaluate the integral, sub $x=y^2$ and get
$$2 \int_0^{\infty} dy \, y^2 \, e^{-y^2} = \int_{-\infty}^{\infty} dy \, y^2 e^{-y^2}$$
Best Answer
First, if $X \sim Exp(\lambda)$, then it has a p.d.f. given by $f_\lambda(x) = \lambda \mathrm{e}^{-\lambda x}$. By the definition of expectation:
$$ \mathrm{E}[Y] = \mathrm{E}[\mathrm{e}^{2X}] = \int_0^\infty \mathrm e^{2x} f_3(x) \mathrm{d} x = 3 \int_0 ^\infty \mathrm{e}^{2x}\mathrm{e}^{ - 3 x}\mathrm d x = 3 \int_0^\infty \mathrm{e}^{-x}\mathrm{d}x = 3.$$