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Precalculus Mathematics Homework Help
Mean, variance and correlation - Multinomial distribution
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[QUOTE="Ray Vickson, post: 5344534, member: 330118"] The event ##\{ X=x, Z=k\}## occurs when there are ##x## 'reds' and ##k-x## 'greens', followed by a 'black', and that has probability [tex] P(X=x,Z=k) = C(k,x)\, p_r^x \, p_g^{k-x} \, p_b [/tex] Thus, ##P(Z=k) = \sum_{x=0}^k P(X=x, Z=k) = (p_r + p_g)^k \, p_b##, by the binomial expansion of ##(p_r + p_g)^k##. Therefore, we have [tex] P(X = x | Z = k) = \frac{P(X=x,Z=k)}{P(Z=k)} = C(k,x)\, \left( \frac{p_r}{p_r+p_g}\right)^x \, \left( \frac{p_g}{p_r+p_g}\right)^{k-x} [/tex]. That means that ##[X|Z=k\ \sim \text{Bin}\,(k, p_r/(p_r+p_g))## as claimed before. Because ##[X|Z=k]## is binomial, its mean is known right away as ##E(X|Z=k) = kp_r/(p_g+p_r)##. Therefore, ##EX = \sum_{k=0}^{\infty} p_b (p_r+p_g)^k k p_r/(p_r+p_g)##. That last summation is do-able. If I were trying to get ##\text{Var}(X)## I would use the fact that ##\text{Var}(X) = E(X^2) - (E X)^2##, then get ##E(X^2)## in a similar manner as ##EX##: ##E(X^2|Z=k)## is the mean of a squared binomial random variable, and that is related to its variance and mean as ##E(X^2|Z=k) = \text{Var}(X|Z=k) + [E(X|Z=k)]^2##. [B]Note added in edit:[/B] Alternatively, you can try to compute the marginal distribution of ##X## directly as [tex] P(X=x) = \sum_{k=x}^{\infty} P(Z=k) P(X=x|Z=k) [/tex] It also turns out that you can almost immediately write down the final expression for ##P(X = x)## without any complicated evaluations, provided that you reason very carefully (and quite subtly) about the nature of the event ##\{ X = x \}##. [/QUOTE]
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Mean, variance and correlation - Multinomial distribution
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