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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.

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  • $\begingroup$ But that's the same thing as the above expression. Doesn't simplify at all. $\endgroup$ – Jerry Oct 30 at 8:41
  • $\begingroup$ What makes you think it can be simplified? $\endgroup$ – Shubham Johri Oct 30 at 8:42
  • $\begingroup$ It should be. I need to compare between different distributions. $\endgroup$ – Jerry Oct 30 at 8:42
  • $\begingroup$ Maybe someone else will be better fit to answer this question. $\endgroup$ – Shubham Johri Oct 30 at 8:43
  • $\begingroup$ Yes, thank you! $\endgroup$ – Jerry Oct 30 at 8:44
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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 $\int_0^1\frac x2dx+1\times1/2=3/4$.

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

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