Probability (Transformation Method)

Question

Let $X$ be a random variable with CDF: $$F_{X}(x)= 1-e^{\lambda x}$$ Let $Y$ be a random variable defined in terms of $X$ as $Y=m$ if $m\leq X < m+1$, where $m$ is a non-negative integer. Compute the distribution of $Y$.

My idea was to apply $$f_{Y}(y)=f_{X}(h^{-1}(y))\left|\frac{d}{dy}h^{-1}(y)\right|$$

First I try $f_{X}(x)= \lambda e^{-\lambda x}$ And the funtion $Y=h(X)=m$. But I can do much with this:$$f_{Y}(y)=f_{X}(h^{-1}(?))\left|\frac{d}{dy}h^{-1}(?)\right|$$

Solution: $Y\sim Geom(1-e^{-\lambda})$ $\Rightarrow$ $f_{Y}(y)=(1-e^{-\lambda})e^{-\lambda}$

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Duplicate of Variables defined as floor and fraction part from exponentially distributed random variable — StubbornAtom, May 27 '20 at 07:26

score 2 · Accepted Answer · edited May 27 '20 at 02:18

To begin with: You probably wanted to write $F_X(x)= \begin{cases} 1-e^{-\lambda x} & x>0 \\ 0 & x \le 0 \end{cases}$, otherwise it isn't good distribution function

You shouldn't use the method to obtain density of $Y$ simply because $Y$ does not have density at all. Note that $Y$ only takes values from $\mathbb N$ so to find its distribution it is enought to find values $\mathbb P(Y=n)$ for any $n \in \mathbb N$. By definition, for $n \in \mathbb N$ we get: $Y=n$ if and only if $X \in [n,n+1)$. So:

$\mathbb P(Y=n) = \mathbb P(X \in [n,n+1)) = \mathbb P(X < n+1) - \mathbb P(X \le n) = e^{-\lambda n}-e^{-\lambda(n+1)} = e^{-\lambda n}(1-e^{-\lambda})$

Writing $p=(1-e^{-\lambda})$, you get $\mathbb P(Y=n) = (1-p)^n p$, so you should recognize (one version of) geometric distribution. So that $Y \sim Geo(1-e^{-\lambda})$

Probability (Transformation Method)

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