- Let $X$ be a random variable with probability function $\operatorname{f}\left(x\right)=\lambda{\rm e}^{-\lambda x}$ if $x>0$ and $\lambda >1$ Is a $constant$. $0$ otherwise.
- Calculate the expected value of the variable $Y=e^X$ by first finding
the density function of $Y$ and applying the elementary definition of
expectation. - As a second method use the unconscious statistician's
theorem.
Attempt
I have already tried, and according to my calculations, the density function of $Y$ is
$$g(y)=\frac{1}{y}\lambda e^{-\lambda (\frac{1}{y})} \text{ if } y>1 (\lambda >1 \text{ constant}),$$
and $0$ otherwise. Are my calculations correct?
Therefore, the expected value is
$$E(Y)=\int_{1}^{\infty}\, y\cdot g(y)\, dy=\int_{1}^{\infty}\, \lambda e^{-\lambda (\frac{1}{y})}\, dy$$
How do I calculate this integral?
Best Answer
The correct Y-density is
$$f_Y(y)=\lambda y^{-(\lambda+1)}$$
$y,\lambda>1$
thus
$$\mathbb{E}[Y]=\int_1^{\infty}\lambda y^{-\lambda}dy=\frac{\lambda}{\lambda-1}$$
LOTUS
$$\mathbb{E}[Y]=\int_0^{\infty}\lambda e^{-\lambda x}e^x dx=\int_0^{\infty}\lambda e^{-(\lambda-1)x}dx=\frac{\lambda}{\lambda-1}\int_0^{\infty}(\lambda-1)e^{-(\lambda-1)x}dx=$$
$$\frac{\lambda}{\lambda-1}\underbrace{\int_0^{\infty}\theta e^{-\theta x}dx}_{=1}=\frac{\lambda}{\lambda-1}$$
Calculation's detail for $f_Y$
$$f_Y(y)=\frac{\lambda}{y}\exp\left\{ -\lambda\log y \right\}=\frac{\lambda}{y}\exp\left\{ \log y^{-\lambda} \right\}=\lambda y^{-\lambda-1}$$