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## Conditional probability mass function of the sum of independent geometric random variables.

$X, Y \sim Geom\left(p\right)$, and $X, Y$ are independent. Find $p_{X | X+Y}\left(i | n\right)$. My approach: pmf of geometric distribution: $p_X(i) = {p(1-p)^{i-1}}$ Then, by the law of total probability: $p_{X | X+Y}\left(i | n\right) = \frac{p_{X, X+Y}\left(i,n\right)}{p_{X+Y}\left(n\right)}$ [Is using the law of total probability here correct as opposed to using Bayes theorem?] […]

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## Conditional variance given condition on both random variables

Correct answer: 2.933 My work: Let $X =$ # of hurricanes hitting Florida, $\sim exp(\lambda = 1.7)$, Y = # of hurricanes hitting Texas, $\sim exp(\lambda = 2.3)$. $$Var(X-Y|X+Y=3) = E[(X-Y)^2|X+Y=3] – E[X-Y|X+Y=3]^2$$ $$=E[X^2|X+=3] – 2E[X|X+Y=3]E[Y|X+Y=3] – E[X-Y|X+Y=3]^2$$ by independence. Since $X$, $Y$ are independent, the joint pmf is simply the product of their […]

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## Conditional variance given condition on both random variables

Correct answer: 2.933 My work: Let X = # of hurricanes hitting Florida, ~exp($\lambda = 1.7$), Y = # of hurricanes hitting Texas, ~exp($\lambda = 2.3$). $Var(X-Y|X+Y=3) = E[(X-Y)^2|X+Y=3] – E[X-Y|X+Y=3]^2 = E[X^2|X+=3] – 2E[X|X+Y=3]E[Y|X+Y=3] – E[X-Y|X+Y=3]^2$ by independence. Since X, Y are independent, the joint pmf is simply the product of their marginal pmfs, […]