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Can you please help me on this question:

(a) In class we define the expectation of a random variable $X$ as $$\mathbb{E}(X):=\sum_{\omega\in\Omega}X(\omega)\mathrm{Pr}[\omega]\;,$$ where $\Omega$ is the sample space. In the notes, the expectation is defined to be $$\mathbb{E}(X):=\sum_{a\in\mathscr{A}}a\mathrm{Pr}[X=a]\;.$$ where $\mathscr{A}$ is the range of values that $X$ can take on. Show that the two definitions are equivalent.

(b) Given a random variable $X$ defined on a sample space $\Omega$, $Y=X^2$ is also a random variable defined on $\Omega$. Why? (Hint: Look review [sic] the definition of a random variable)

(c) Show, from the definition of the expectation, that $$\mathbb{E}(X^2)=\sum_{a\in\mathscr{A}}a^2\operatorname{Pr}[X=a].$$ (You can start with the definition in class or the definition in the notes, since you have already shown in part (a) that they are equivalent.)

(d) Generalize the result in part (c) to give an expression for $\mathbb{E}(f(X))$, where $f$ is an arbitrary function from $\mathfrak{R}$ to $\mathfrak{R}$. You can give your answer without proof.

Thanks!

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It is not acceptable to link to an external file containing a picture of your homework sheet. –  Phira Oct 27 '11 at 23:30
    
Sometimes it is. But not without asking a concrete question about it, and/or showing some significant work on it. –  Henning Makholm Oct 27 '11 at 23:31
    
well i know how to approach the question...its just i don't know how to write a formal proof –  LiveYourLife Oct 27 '11 at 23:32
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Fascinating. Tell us more! (Preferably by editing the question and adding your explanation there). –  Henning Makholm Oct 27 '11 at 23:34
    
@HenningMakholm I know, the point is, it is not acceptable to link to an external picture as an entire question. –  Phira Oct 27 '11 at 23:38
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2 Answers 2

Here's an almost complete solution to (a) with some justifications left out. $$\begin{align*} E(X) & = \sum_{a\in\mathscr{A}} a\cdot \text{Pr}[X=a]\\ & = \sum_{a\in\mathscr{A}} ~~a\left(\sum_{\omega\in[X=a]} \text{Pr}[\omega]\right)\\ & = \sum_{a\in\mathscr{A}} ~~~~\sum_{\omega\in[X=a]} a\cdot \text{Pr}[\omega]\\ & = \sum_{a\in\mathscr{A}}~~~~~ \sum_{\omega\in[X=a]} X(\omega)\text{Pr}[a]\\ & = \sum_{\omega\in\Omega} X(\omega)\text{Pr}[\omega] \end{align*} $$ where the last equality is true since the events $[X=a]~\forall a\in \mathscr{A}$ partions the sample space, $\Omega$. Thus, summing over all the outcomes is the same as summing over $\Omega$.

I leave the justification of the other equalities for you.

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Thanks Nana! I was wondering if you could explain how to get from step 3 to step 4 please? –  LiveYourLife Oct 28 '11 at 1:22
    
By steps 3 and 4, I hope you mean third and fourth equalities. Well, I used the definition of the event $[X=a]$. –  Nana Oct 28 '11 at 1:54
    
Okay, thanks for the clarification! I think I understand this question now. For part b) to prove that X^2 is also a random variable, we can assume each of the X that get multiplied are random variables...how do we show that "randomness" is preserved through multiplication? –  LiveYourLife Oct 28 '11 at 2:07
    
Part(b) just uses the definition of a random variable: a function from a sample space $\Omega$ into the real numbers. –  Nana Oct 28 '11 at 2:38
    
can you elaborate on that? part c uses b and some kind of proof.. –  LiveYourLife Oct 28 '11 at 2:41
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For (a): For each $a\in\mathscr{A}$ let $\Omega_a=\{\omega\in\Omega:X(\omega)=a\}$. How is $$\sum_{\omega\in\Omega_a}X(\omega)\mathrm{Pr}[\omega]$$ related to $a\mathrm{Pr}[X=a]$?

Added: For each $\omega \in \Omega_a$, $X(\omega)=a$, so $$\sum_{\omega\in\Omega_a}X(\omega)\mathrm{Pr}[\omega]=a\sum_{\omega\in\Omega}\mathrm{Pr}[\omega]\;;$$ if you can show that $$\sum_{\omega\in\Omega_a}\mathrm{Pr}[\omega] = \mathrm{Pr}[X=a]\;,$$ you’ll be done. Now, how do you define $\mathrm{Pr}[X=a]$?

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We know that $\{\omega\in\Omega: X(\omega) = a\}$ –  LiveYourLife Oct 27 '11 at 23:49
    
hmm...i am really confused as to how to approach this question. any help? –  LiveYourLife Oct 27 '11 at 23:56
    
@LiveYourLife: Sorry: I meant to write $\Omega_a=\{\omega\in\Omega:X(\omega)=a\}$. I’ll fix it and add a little more. –  Brian M. Scott Oct 28 '11 at 1:16
    
Okay, thanks for the clarification! I think I understand this question now. For part b) to prove that X^2 is also a random variable, we can assume each of the X that get multiplied are random variables...how do we show that "randomness" is preserved through multiplication? –  LiveYourLife Oct 28 '11 at 1:28
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