$X \sim U(0,1)$ be uniformly distributed on $(0,1)$ and $Y \equiv -\frac{1}{\lambda} \log X$. Compute the distribution of $Y$.

Question

Let $X \sim U(0,1)$ be uniformly distributed on $(0,1)$ and let $Y \equiv -\frac{1}{\lambda} \log X$ and $\lambda > 0$.

I want to compute the distribution of $Y$.

Attempt:

I tried to use the "change-of-variable"-method.

Let $\mathbb{E}[h(Y)] = \mathbb{E}[h(-\frac{1}{\lambda} \log X)] = \int_0^1 h(-\frac{1}{\lambda} \log x) dx$.

Now substituting $u= -\frac{1}{\lambda} \log x \iff x = e^{-u\lambda}$ and $dx = -du \lambda x$

$\implies \int_0^1 -h(u)du\lambda e^{-u \lambda}$.

Now as I see it the distribution should now be $-\lambda e^{-u\lambda}$.

But the distribution should be exponential hence $\lambda e^{-u \lambda}. $

I'm a bit confused. The minus sign does matter in terms of distribution, right?And is it correct that in this case distribution as well as density are the same?

Answer 1

You forgot to change the bounds. When $x\to0$ you have $u\to+\infty$ and when $x=1$ you have $u=0$, hence $$ \int_0^1h\left(-\frac1\lambda\log x\right)\,dx=\int_{+\infty}^0-h(u)\lambda\exp(-u\lambda)\,du=\int_0^{+\infty}h(u)\lambda\exp(-u\lambda)\,du. $$ and now you find a meaningful distribution.

Answer 2

You did make the substitution correctly.

$\int_0^{1} g(x)dx=\int_0^{\infty} g(e^{-\lambda u}) \lambda e^{-\lambda u} du$.