1
$\begingroup$

In my assignment, I've came up across this question:

Let $X$ be a discrete random variable which attains the values $x_1 < \dots < x_n$ with probabilities $p_1, \dots , p_n$. With $F$ the cdf of $X$, what is the cdf of $F(X)$?

I know that $F(X)$ is: $$0;(-\infty,x_1)\\ p_1;[x_1,x_2)\\ \vdots\\ p_1+\dots+p_{n-1};[x_{n-1},x_n)\\ 1;[x_n,\infty)$$

Now, from this I am supposed to create another CDF. Is this even possible? I found a solution that goes like this:

$$0;(-\infty,0)\\ \frac{x_1-m}{p-m};[0,p_1)\\ \vdots\\ \frac{x_n-m}{p-m};[p_{n-1},p_n)\\ 1;[p_n,\infty)$$ With $m$ going to minus infinity and $p$ going to plus infinity. I cannot wrap my head around the $p$ and $m$ in the fractions, are they supposed to be "normalizing" the distribution function so it goes to 1? How does it work?

$\endgroup$
1
$\begingroup$

Firstly note that $F(x)=\mathbb P(X\leq x)$ is the function of $x$, not of $X$: $$ F(x) = \begin{cases}0, & x\in (-\infty, x_1), \cr p_1, & x\in[x_1,x_2),\cr \dots & \cr p_1+\dots+p_{n-1}, & x\in[x_{n-1},x_n),\cr 1, & x \in [x_n,\infty). \end{cases} $$

Let $Y=F(X)$. You need to substitute $X$ in the function $F(x)$ instead of variable. Since $X$ can take only values $x_1,\ldots,x_n$, $F(X)$ is also discrete random variable. What are its values?

In the case when $X=x_1$, $Y=F(X)=F(x_1)=p_1$. So $\mathbb P(Y=p_1)=\mathbb P(X=x_1)=p_1$.

If $X=x_2$, then $Y=F(X)=F(x_2)=p_1+p_2$. And $\mathbb P(Y=p_1+p_2)=\mathbb P(X=x_2)=p_2$.

Continue for all values of $X$. And then construct CDF of $Y$.

The solution that you found somewhere does not apply to this task.

$\endgroup$
2
  • $\begingroup$ Hence, the CDF of $F(Y)$ is going to look like: $$ F(y)= \begin{cases} 0, & y \in (-\infty,p_1)\\ p_1, & y \in [p_1,p_1+p_2)\\ \vdots\\ p1+\dots+p_{n-1} & y\in [p_1+\dots +p_{n-1},p_1+\dots +p_{n})\\ 1 & y \in [p_1+\dots +p_{n},\infty) \end{cases} $$ $\endgroup$ – PK1998 Dec 21 '19 at 17:58
  • $\begingroup$ @PK1998 Yes, this is right. $\endgroup$ – NCh Dec 22 '19 at 4:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.