2
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I am given $$Q \sim\mathrm{Beta} (2,3)$$ $$[X|Q=q]\sim\mathrm{Bin} (3,q)$$

I am asked to find (1) $E[X]$, (2) $\operatorname{Var}[X]$, (3) $E[Q|X]$, (4) $\operatorname{Var}[Q|X]$, (5) $v(q|X=x)$.

  1. I understand $E[X]=E[E[X|Q]]=E[xQ]=\frac{x\alpha}{\alpha + \beta}$
  2. $\operatorname{Var} [X]=E[\operatorname{Var}[X|Q]]+\operatorname{Var}[E[X|Q]]=E[xQ(1-Q)]+\operatorname{Var}[xQ]$
  3. No idea
  4. No idea
  5. I know I can compute $\frac{p(x,q)}{f(x)}$ but can I say $f(x)\sim\mathrm{Bin} (3,q)$ given our assumptions?
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  • $\begingroup$ Have you checked en.wikipedia.org/wiki/Beta-binomial_distribution ? $\endgroup$ – Tim Nov 7 '16 at 19:51
  • $\begingroup$ Hey @Tim . Yes, I have. The article mentions briefly $P[X|Q=q]$ but never the reverse, $P[Q|X=x]$. Is this a straight computation or is there a simple expression to describe $E[Q|X]$ and $Var[Q|X]$? $\endgroup$ – ozarka Nov 7 '16 at 20:13
  • $\begingroup$ Hints: 3 and 4 can be found using Bayes' Rule. Don't worry about $f(x)$, just write out the expression and see if you can recognize the distribution for $Q$ from the function. If it's hard, write it out again, leaving out everything that doesn't have to have a $Q$ in it (e.g., $x\exp \{Q/x\}$ can have the initial $x$ removed) and try again. $\endgroup$ – jbowman Nov 7 '16 at 20:40
  • $\begingroup$ @jbowman, I think I got it. (3) is $\frac{a}{a+b}=\frac{x+2}{x+2+6-x}$. (4) is $\frac{ab}{(a+b)^{2}(a+b+1)}$ using $a=x+2$ and $b=6-x$. For (5), I know that $\frac{p(x,q)}{f(x)}=\frac{g(q) u(x|q)}{P[X=x]}$. Is $P[X=x]$ also follow the Binomial distribution $Bin(3,q)$? $\endgroup$ – ozarka Nov 7 '16 at 21:18
  • $\begingroup$ Very good! But the unconditional distribution of $x$ is not binomial. What is question 5? You don't have $v$ defined anywhere... $\endgroup$ – jbowman Nov 7 '16 at 21:38

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