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I was wondering if there is a shortcut or an easy way to calculate the expectation of a certain Random variable which is equal to two random variables with different probabilities.

For example: We are given random variables $X[1]$ and $X[2]$ (the numbers in the brackets are indexes) which have normal Gaussian distributions with $E(X[1])=E(X[2])=1$, $Var(X[1])=Var(X[2])=1$, and R.V $Y[1]$ and $Y[2]$ which have exponential distributions with parameter $lambda =2$, such that all of the R.V $X[1],X[2],Y[1]$ and $Y[2]$ are independent. If we define new random variables $Z[1]$ and $Z[2]$ such that $Z[i]$ equals $X[i]$ with probability $1/3$ and equals $Y[i]$ with probability $2/3$ for $i=1,2$, what is, for example, the expectation of $E(Z[1])$ and $E(Z[1]\cdot Z[2])$?

I tried finding the probability distribution of both $Z[1]\cdot Z[2]$ and $Z[1]$ and tried to calculate the respective expectations by integrating which is a rather long and gruesome procedure.

Thanks for any help in advance.

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  • $\begingroup$ I assume in my answer that $Z[1]$ and $Z[2]$ are independent. This is not actually specified. Did you mean to put this in the question? $\endgroup$ – Matt Samuel Oct 4 '15 at 12:02
  • $\begingroup$ Thanks a lot for the answer for the first expectation, and no, I missed one detail: It is given that at any moment, if Z[1] equals X[1] then Z[2] also equals X[2] for sure, and the same goes with Y[1] and Y[2]. In other words, it's like at each moment a robot picks either the vector (X[1], X[2]) with probability 1/3 and assigns it to the vector (Z[1], Z[2]) or it picks the vector (Y[1], Y[2]). $\endgroup$ – Dylan132 Oct 4 '15 at 12:05
  • $\begingroup$ Then I'll fix the expectation of the product. $\endgroup$ – Matt Samuel Oct 4 '15 at 12:21
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By the law of total expectation we have $$E[Z[i]]=E[E[Z[i]|X[i],Y[i]]]=E[\frac{1}{3}X[i]+\frac{2}{3}Y[i]]=\frac{1}{3}E[X[i]]+\frac{2}{3}E[Y[i]]$$ For the product, $$E[Z[1]Z[2]]=E[E[Z[1]Z[2]|X[1],Y[1],X[2],Y[2]]]=E[\frac{1}{3}X[1]X[2]+\frac{2}{3}Y[1]Y[2]]=\frac{1}{3}E[X[1]]E[X[2]]+\frac{2}{3}E[Y[1]]E[Y[2]]$$ This should solve your problem.

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  • $\begingroup$ Thanks a lot Sir! I'm glad you told me about the total law of expectation, never even knew it existed...! $\endgroup$ – Dylan132 Oct 4 '15 at 12:38
  • $\begingroup$ No problem. @dylan132 $\endgroup$ – Matt Samuel Oct 4 '15 at 12:40

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