[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:

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

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