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?


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.

  • $\begingroup$ 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

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