# Integration with distribution function [closed]

What is the next step for $$\int_{0}^{1} F(x)dx$$ given I don't have an exact expression for $$F(x)$$?

Edit:

This above expression came after integrating by parts the following:

$$F(x)=\begin{cases}0,&x<0\\x/2,&0\le x<1\\1,&x\ge1\end{cases}$$

I'm just trying to find the expected value (PDF?) of the function.

• But that's the same thing as the above expression. Doesn't simplify at all. – Jerry Oct 30 at 8:41
• What makes you think it can be simplified? – Shubham Johri Oct 30 at 8:42
• It should be. I need to compare between different distributions. – Jerry Oct 30 at 8:42
• Maybe someone else will be better fit to answer this question. – Shubham Johri Oct 30 at 8:43
• Yes, thank you! – Jerry Oct 30 at 8:44

I presume the question is to find the expected value of the random variable with the given cumulative distribution function. The distribution function is a staircase function with finite jumps at $$2$$ points, namely $$0,1$$, so the random variable $$X$$ must be discrete and takes only two values $$0,1$$ (Bernoulli). So$$P(X=0)=F(0)-F(0^-)=1/2\\P(X=1)=F(1)-F(1^-)=1/2\\\Bbb E[X]=0\times1/2+1\times1/2=1/2$$

Edit:

The distribution function is still discontinuous at $$1$$ but non-constant in $$[0,1)$$, so it seems like your random variable is a mix of a continuous and discrete random variable.

The relation $$P(X=1)=F(1)-F(1^-)=1/2$$ remains intact.

In $$[0,1)$$, the PDF is given by $$F'(x)=1/2$$.

So the required expectation is $$\int_0^1\frac x2dx+1\times1/2=3/4$$.

• Thanks, Shubham :) I think it takes continuous, not discrete values though :( – Jerry Oct 30 at 9:23
• @Gerard I think you should recheck the distribution function. In the comments you wrote $x/2$ but in your edit you wrote $1/2$ when $0\le x<1$. – Shubham Johri Oct 30 at 9:24
• Oops, yes, it's x/2 not 1/2 – Jerry Oct 30 at 9:25
• @Gerard Look at the edit. – Shubham Johri Oct 30 at 9:41
• @Jerry Please accept the answer if you have no questions anymore. – callculus Oct 31 at 20:24