# Thread: Functions of one random variable question

1. ## Functions of one random variable question

I have a problem as follows:
Let X have the pdf f(x)= θ * e( - θ x), 0 < x < infinity

find the pdf of Y = ex

I have
G(y) = P(X < ln y) = integral of
θ * e( - θ x) dx from 0 to ln y = 1 - y-θ

So then I need g(y) = G'(y) of Y? which would be
θ * e( - θ (ln y)??

or am I missing something super-duper obvious?

Thanks

2. ## Re: Functions of one random variable question

Note that the random variable $X \sim \mbox{exp}(\theta)$ (exponential distribution) where the distribution function $F$ is given by $F_{X}(x) = 1-e^{-\theta x}$ (for $x \geq 0$). We can compute the distribution function of the random variable $Y=e^X$ as follows: $F_{Y}(x) = P(Y \leq x) = P(e^X \leq x) = P(X \leq \ln(x)) = F_{X}[\ln(x)]$ therefore
$F_{Y}(x) = 1-e^{-\theta \ln(x)}$. We can find the pdf of $Y$ (let's call it $g(x)$) by calculating the first derivative of the distribution function, hence $g(x) = \frac{dF_{Y}(x)}{dx} = \frac{d}{dx}\left[1-e^{-\theta \ln(x)}\right] = \left(\frac{\theta}{x}\right)e^{\theta \ln(x)}$