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[Not Homework]

Hey guys, I've been having trouble with this concept for a while as I just cant seem to get it down, can anyone help me with regards to this question I found in a paper?

Question

My understanding: The expectation of a RV is just the probability of it happening, Hence,

for i) $$ E[X_i] = P(X_i = 1) = P(r_i\hspace{0.1cm}is\hspace{0.1cm}withdrawn) $$

However, since the balls are drawn without replacement, I'm unsure how as to get this probability (as ri would be first, or second etc) Hence, is it:

$$ \frac{1*\binom{29}{11}}{\binom{30}{12}} $$

where 1 is fixing ball ri, and the rest is picking any balls at random?

for ii) and iii), I'm pretty much stumped. Can I assume their independence and thus,

$$ E[X_iY_i] = E[X_i]*E[Y_i] $$

Thank you for your help and guidance! :D

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Just to be clear: The expectation of an Indicator Random Variable is the probability of it being 1.

$$\begin{align}\mathsf E(X_i) & = 0\cdot \mathsf P(X_i=0)+1\cdot \mathsf P(X_i=1)\\ & = \mathsf P(X_i=1)\end{align}$$


$X_i$ is the indicator that red ball #$i$, for $i\in\{1..10\}$, is one of the $12$ out of $30$ balls drawn.

Imagine we lay the balls in a row and say the first $12$ balls will be the ones drawn.   Of all the ways to arrange the balls, out red ball #$i$ will be among the favoured set $12$ in every $30$ arrangements.

Thus $\mathsf P(X_i=1)=\frac{12}{30}$

Thus $\mathsf E(X) $ $=\sum_{i=1}^{10} \frac {12}{30} \\= 4$

One of the things new students often find counter-intuitive is that Linearity of Expectation does not require the summed random variables to be independent.

Similarly $\mathsf E(Y) $ $= \sum_{j=1}^8 \mathsf E(Y_j)\\= 3.2$


However, when it comes the the expectation of multiples of random variables, you do need to consider their joint probability, because $$\begin{align}\mathsf E(XY) & = \mathsf E((\sum_{i=1}^{12}X_i) (\sum_{j=1}^8 Y_j)) \\ & = \sum_{i=1}^{12}\sum_{j=1}^8\mathsf P(X_i=1,Y_j=1)\end{align}$$

Can you complete?

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  • $\begingroup$ Thanks for your answer! :) If Xi is the indicator for ball i, going by your definition, then wouldnt the probability of Xi be 1/12 instead? Similarly for Yi.. $\endgroup$ – Wboy Nov 22 '15 at 14:39

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