expected value of an exponential distribution

Question

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.

Answer 1

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:

$$ \mathrm{E}[Y] = \mathrm{E}[\mathrm{e}^{2X}] = \int_0^\infty \mathrm e^{2x} f_3(x) \mathrm{d} x = 3 \int_0 ^\infty \mathrm{e}^{2x}\mathrm{e}^{ - 3 x}\mathrm d x = 3 \int_0^\infty \mathrm{e}^{-x}\mathrm{d}x = 3.$$

Answer 2

Let $X\sim\mathsf{Expo}(\lambda)$ and $Y=e^{aX}$ with $\lambda>a>0$. Then for $y>1$ we have \begin{align} \mathbb P(Y\leqslant y) &= \mathbb P\left(e^{aX}\leqslant y\right)\\ &= \mathbb P\left(X\leqslant \frac{\log y}a\right)\\ &= 1 - e^{-\frac{\lambda}a\log y}\\ &= 1 - \left(e^{\log y}\right)^{-\frac\lambda a}\\ &= 1 - y^{-\frac\lambda a}. \end{align} It follows that \begin{align} \mathbb E[Y] &= \int_1^\infty \left(1-\mathbb P(Y\leqslant y)\right)\ \mathsf dy\\ &= \int_1^\infty y^{-\frac\lambda a}\ \mathsf dy\\ &= \frac\lambda{\lambda-a}. \end{align} If $\lambda=3$ and $a=2$ then this is equal to $3$.