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I am working on this question Conditional probability - need help on calculating numerator and I got into use of the law of total expectation. How using this law I can find $$ E(X_1^2X_2^4X_3^6\mid X_1+X_2+X_3=A), $$ where $X_i$ are geometric random variables representing the number of failures with parameter $p$.

I am not sure that I am using the law properly. I am starting with $E(E(E(X_3^6\mid A-X_1-X_2=a_3)\times X_2^4\mid A-X_1=a_2)\times X_1^2\mid X_1+X_2+X_3=A)$

Michael Hardy
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Forbs
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1 Answers1

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One way to go is to figure out the joint distribution of $(X_1,X_2,X_3)$ given $X_1+X_2+X_3=A$. Take $(x_1,x_2,x_3)$ so that $x_1,x_2,x_3 \geq 0$ and $x_1+x_2+x_3 = A$. Try computing $$ P(X_1=x_1,X_2=x_2,X_3=x_3 | X_1+X_2+X_3=A) $$ by using the definition of conditional expectation. See here for a similar calculation where you have two instead of three random variables. I think you should get that $(X_1,X_2,X_3)$ given $X_1+X_2+X_3=A$ is uniformly distributed, where each possible value $(x_1,x_2,x_3)$ with $x_1+x_2+x_3=A$ has probability $\frac2{(A+1)(A+2)}$. Then $$ E[X_1^2X_2^4X_3^6 | X_1+X_2+X_3=A] = \frac2{(A+1)(A+2)} \sum_{(x_1,x_2,x_3): x_1+x_2+x_3=A} x_1^2 x_2^4 x_3^6. $$ This at least gives you a computable formula for the conditional expectation.

spalein
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  • Thank you. May I ask you to elaborate on why each $(x_1,x_2, x_3)$ has probability $2/(A+1)(A+2)$? – Forbs Jul 16 '20 at 19:54
  • Calculate $P(X_1=x_1,X_2=x_2,X_3=x_3|X_1+X_2+X_3=A)$. This probability is only nonzero when $(x_1,x_2,x_3)$ satisfies $x_1,x_2,x_3 \geq 0$ and $x_1+x_2+x_3=A$. For such $(x_1,x_2,x_3)$, you have \begin{align} & P(X_1=x_1,X_2=x_2,X_3=x_3|X_1+X_2+X_3=A) \ & =P(X_1=x_1,X_2=x_2,X_3=x_3)/P(X_1+X_2+X_3=A). \end{align} You can compute the numerator by using that $X_1$, $X_2$ and $X_3$ are iid geometric, and you can compute the denominator by using that $X_1+X_2+X_3$ has the negative binomial distribution. You should end up with $2/((A+1)(A+2))$. Try working it out, happy to help if you get stuck. – spalein Jul 16 '20 at 20:58
  • Thank you for your help. – Forbs Jul 16 '20 at 21:36