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A random variable X has the cumulative distribution function:

$F(x)= \left\{ \begin{array}{l} 0 \text{ for x < 1}\\ \cfrac{x^2-2x+2}{2} \text{ for } 1 \le x <2\\ 1 \text{ for } x \ge 2 \end{array} \right. $

Calculate the variance of x.

I know the definition of variance is: $Var[X]=E[X^2]-E[X]^2$ so we proceed by calculating each term.

$E[X]=\int\limits_{-\infty}^{\infty}xf(x)dx\tag{1}$

$E[X^2]=\int\limits_{-\infty}^{\infty}x^2f(x)dx\tag{2}$

So to evaluate equations (1) and (2), we need to calculate the PDF. This is where I am confused. I know the PDF is the derivative of the CDF. Also, we note that $F(2)-F(1) = 1/2$ and by definition $F(\infty)=1$.

This forces the probability for $x<1$ to be 1/2. Is this correct? How would we write the system of equations for the PDF below? I don't really see how this is a combination of discrete and continuous if the limits are $(x<1), 1 \le x < 2, x \ge 2$. Can someone please explain this and breakdown $E[X],E[X^2]$?

$f(x)= \left\{ \begin{array}{l} 0 \text{ for x < 1}\\ (x-1) \text{ for } 1 \le x <2\\ 0 \text{ for } x \ge 2 \end{array} \right. $

Any help is appreciated. Thank you!

user1527227
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1 Answers1

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The CDF as given explicitly states that Pr{ X < 1 } = 0, so the answer to your first question is NO, it is not ½.

However, the CDF jumps from 0 to ½ at X = 1. Therefore we know that Pr{ X = 1 } = ½.

I think this will be enough to get you started on the answer. I could work it all out for you, but it looks like a homework problem so I think I will stop here.

LorenCobb
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