Determine the distribution of $Y$

Question

Let $X$ be continuously uniformly distributed on $(0, 1)$. Let $\lambda> 0$ and $Y = -\frac{1}{\lambda}\ln (1 - X)$.

Determine the distribution of $Y$. Make a note of the intermediate steps.

That means that we have to calculate the distribution function of $Y$, i.e. $F_Y(y)$, right?

We have the following :

$$\begin{align*}F_Y(y)&=P(Y\leq y)\\&=P\left (-\frac{1}{\lambda}\ln (1 - X)\leq y\right )\\&=P\left (-\ln (1 - X)\leq \lambda y\right )\\&=P\left (\ln (1 - X)\geq -\lambda y\right )\\ & =P\left (1 - X\geq e^{-\lambda y}\right )\\&=P\left ( X\leq 1-e^{-\lambda y}\right )\\&=F_X(1-e^{-\lambda y})\end{align*} $$

Is that correct and complete?

Answer 1

If $X \sim U(0,1)$, then $F_X(x)=x$ for $0 < x \leq 1$. Then,

$$F_Y(y)=F_X(1-e^{-\lambda y})=1-e^{-\lambda y}.$$

What does that mean? $Y \sim ??$

Answer 2

A separate approach.

We want to transform $x\mapsto -\frac{1}{\lambda}\ln(1-x)$ .

Then you have the $\frac{dy}{dx}=\frac{1}{\lambda(1-x)}>0$ for $x\in(0,1)$.

So you have $f_{Y}(y)=|\frac{dx}{dy}|f_{X}(x)=\lambda(1-x)=\lambda e^{-\lambda y}\,\,,y>0$.

So the pdf of $Y$ is that of a $\text{Exp}(\lambda)$ variate ..

So cdf of $Y$ is $F_{Y}(y)=\begin{cases}1-e^{-\lambda y} \, ,y\geq 0\\ 0 \,,\text{elsewhere}\end{cases} $

This is what is known as the Jacobian Transformation.

In general :-

If you want to transform $X$ to $Y(X)$.

Then the pdf $f_{Y}(y)=|\frac{dx}{dy}|f_{X}(x)$ . ( In the RHS you just replace x with the expression of x in terms of $y$, i,e the inverse transformation).

This can be generalized to higher dimensions.

If you want to transform $(X_{1},X_{2},...X_{n})$ to $(U_{1},U_{2},..,U_{n})$ where $U_{i}$'s are functions of $(X_{1},X_{2},...X_{n})$ Then $$f(u_{1},u_{2},...u_{n})=|\frac{\partial(x_{1},x_{2},..x_{n})}{\partial(u_{1},u_{2},..,u_{n})}|f(x_{1},x_{2}...,x_{n})$$