# CDF of a cumulative distribution function of a discrete random variable

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.

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