# variance of the cdf with a jump

The problem is

A random variable X has the cdf $$F(x)=\frac{x^2-2x+2}{2}\quad\text{if}\quad1\leq x<2$$

and $F(x)=0$ when $x<1$, $F(x)=1$ when $x\geq 2$.

Calculate the variance of X(the answer is $\frac{5}{36}$)

My question is

1. What is the relationship between with the cdf and pdf when cdf has a jump at the point x=1;

2. What is the definition of $E[X^2]$ when pdf is not continuous at the point x=1?

I think this is a case for the r.v. is partially discrete and partially continuous, so, I don't know the definition. Maybe it needs some knowledge of lebesgue stieltjes integral, I'm not sure. Could you please help me? Thank you so much!

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With a distrete random variable the cumulative distribution function has jumps all the way where each step represents a probability; with a continuous random variable the cumulative distribution function is continuous. This example is just a combination of the two.

You can find $\Pr(X=1)=1/2$ and that with $1 \lt x \lt 2$ the density is $f(x)=x-1$. Just treat the discrete and continuous parts as you would usually.

So you have $E[X]= 1 \times \frac{1}{2} + \int_1^2 x(x-1) dx = \frac{4}{3}$

and similarly $E[X^2]= 1^2 \times \frac{1}{2} + \int_1^2 x^2(x-1) dx = \frac{23}{12}$

so the variance is $\frac{23}{12} - \left(\frac{4}{3}\right)^2 = \frac{5}{36}.$

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Ok, got it! Thank you so much! –  breezeintopl Jan 1 '12 at 19:50

$F(1)= \frac{1}{2}$. So the density function is given by $$f(x)=\begin{cases}\frac{1}{2} & x=1\\ x-1 & 1\lt x \lt 2\\ 0 & \text{elsewhere} \end{cases}$$ Thus, we have $$E(X) = \frac{1}{2} + \int_1^2 x(x-1)~\text{d}x$$ $$E(X^2) = \frac{1}{2} + \int_1^2 x^2(x-1)~\text{d}x$$ from which you can find $Var(X)$.

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It is diffioult to say $f(1)=\frac{1}{2}$ is the density here, as it is a point probability. –  Henry Jan 1 '12 at 19:52
Thank you so much! –  breezeintopl Jan 1 '12 at 19:53