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This is a sample question in the book, and I already have the answer, there is one step I am really confused about:

$$X=0.5X_1+0.5X_2$$ So I know: $$Var[X]=E[X^2]−E[X]^2$$ and then I already figured out $E[X]^2$, $E((X_1)^2)$, and $E((X_2)^2)$ However, for $E[X^2]$, as indicated in the book, it says: $$E(X^2)=\frac{1}{2}E((X_1)^2)+\frac{1}{2}E((X_2)^2)$$ Where did this come from?

The originally question is that given the CDF of the function as: $$F_x(x)=\left\{\begin{matrix} 0, for x<1\\ \frac{x^2-2X+2}{2}, for 1\leq x< 2\\ 1, for x\geq 2 \end{matrix}\right.$$

I managed to figure out that $X_1$ is a discrete function with probability of 1 at x=1, and everything else 0. The continuous function $X_2$ has probability $2x-2$ for $1<x<2$, and 0 elsewhere.

Any help is appreciated. Thanks in advance.

ttothef
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  • Does the book say anywhere that $X_1$ and $X_2$ are independent or uncorrelated? –  Feb 20 '14 at 00:21
  • I guess it is independent, the originally question is that: http://math.stackexchange.com/questions/544690/variance-for-mixed-distribution-continuous-discrete?rq=1 this link has the originally question, It was originally given in CDF of the mixture of two distributions, so I guess it is independent – ttothef Feb 20 '14 at 00:38

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To find the expectation of $(0.5X_1+0.5X_2)^2$, expand the square and use the linearity of expectation. We get $$(0.5)^2E(X_1^2)+(0.5)^2E(X_2^2)+2(0.5)^2E(X_1X_2).$$ But in our case, $X_1X_2$ is always $0$, so has expectation $0$. Please note that the constants in the formula of the OP are not correct.

André Nicolas
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  • First, why does Expectation of $X_1X_2$ 0? And which constants is not correct? are you suggesting that it should be 0.5^2$E(X^2)$ instead of 1/2$E(X1^2)$? Thanks a lot! – ttothef Feb 20 '14 at 02:08
  • Let us find the probability that $X_1X^2\gt a\gt 0$. For $x\gt a$, we have $\Pr(X_1\ge x)=0$. More informally, $X_1X_2$ cannot ever have value other than $0$. And yes, I do mean $(0.5)^2$ instead of $\frac{1}{2}$. – André Nicolas Feb 20 '14 at 02:29