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If $f(x_1,x_2)= {x_1 \choose x_2}(1/2)^{x_1}(x_1/15)$ where $x_2 = 0,1,\ldots,x_1$ and $x_1 = 1,\ldots,5$, how would you find $E[X_2]$ and $E[X_2|X_1=x_1]$


To find $E[X_2]$ you just perform a double summation from $0$ to $x_1$ for $x_2$ and from $1$ to $5$ for $x_1$, multiplying each value of $x_2$ by the pdf.

To find $E[X_2|X_1=x_1]$, you fix $x_1$, and add the products of all relevant $x_2$ by the pdf.

So for example: if $x_1 = 1$, then $E[X_2|X_1= 1]= 0*pdf + 1* pdf = {1 \choose 1} (1/2)(1/15)= 1/30$. Is the correct reasoning? Do I repeat this process for all the other $x_1$? I'm trying to find an easier way to do this problem.

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up vote 1 down vote accepted

By the tower property, $E[X_{2}] = E[ E[X_{2} | X_{1} = x_{1}] ]$, so I would first start by solving for $E[X_{2} | X_{1} = x_{1}]$: $$E[X_{2}|X_{1} = x_{1}] = \sum_{x_{2}=0}^{x_{1}}x_{2}\cdot{}P(X_{2}=x_{2}|X_{1}=x_{1}) = \sum_{x_{2}=0}^{x_{1}}x_{2}\cdot{}\frac{P(X_{2}=x_{2} \cap X_{1}=x_{1})}{P(X_{1}=x_{1})} $$

$$ = \sum_{x_{2}=0}^{x_{1}}x_{2}\cdot{}\frac{\biggl[\begin{pmatrix} x_{1} \\ x_{2}\end{pmatrix} \biggl(\frac{1}{2}\biggr)^{x_{1}}\biggl( \frac{x_{1}}{15}\biggr)\biggr]}{\displaystyle\sum_{x_{2}=0}^{x_{1}}\biggl[\begin{pmatrix} x_{1} \\ x_{2}\end{pmatrix} \biggl(\frac{1}{2}\biggr)^{x_{1}}\biggl( \frac{x_{1}}{15}\biggr)\biggr]}.$$

Next, write down the marginal probability mass function of $X_{1}$, which is given by summing out the $x_{2}$ variable, which is the numerator above.

$$ P(X_{1} = x_{1}) = \sum_{x_{2}=0}^{x_{1}}\biggl[\begin{pmatrix} x_{1} \\ x_{2}\end{pmatrix} \biggl(\frac{1}{2}\biggr)^{x_{1}}\biggl( \frac{x_{1}}{15}\biggr)\biggr].$$

Now you can compute $E[X_{2}]$ as the expectation of $E[X_{2}|X_{1}=x_{1}]$ using the PMF of $X_{1}$: $$ E[X_{2}] = \sum_{x_{1}=1}^{5}\Biggl[\sum_{x_{2}=0}^{x_{1}}x_{2}\cdot{}\frac{\biggl[\begin{pmatrix} x_{1} \\ x_{2}\end{pmatrix} \biggl(\frac{1}{2}\biggr)^{x_{1}}\biggl( \frac{x_{1}}{15}\biggr)\biggr]}{\displaystyle\sum_{x_{2}=0}^{x_{1}}\biggl[\begin{pmatrix} x_{1} \\ x_{2}\end{pmatrix} \biggl(\frac{1}{2}\biggr)^{x_{1}}\biggl( \frac{x_{1}}{15}\biggr)\biggr]}\cdot{}\sum_{x_{2}=0}^{x_{1}}\biggl[\begin{pmatrix} x_{1} \\ x_{2}\end{pmatrix} \biggl(\frac{1}{2}\biggr)^{x_{1}}\biggl( \frac{x_{1}}{15}\biggr)\biggr]\Biggr] $$

$$ = \sum_{x_{1}=1}^{5}\sum_{x_{2}=0}^{x_{1}}x_{2}\cdot{}\biggl[\begin{pmatrix} x_{1} \\ x_{2}\end{pmatrix} \biggl(\frac{1}{2}\biggr)^{x_{1}}\biggl( \frac{x_{1}}{15}\biggr)\biggr]$$

which is just the basic double sum formula for $E[X_{2}]$ as expected.

As for what these sums work out to, I would probably just use SymPy, Maple, or Mathematica if you want closed forms, or NumPy or Matlab if you want the numerical results.

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I think E[X2] eventually equals the first term in the product – lord12 Mar 20 '12 at 0:53
But the first term of the product is just for a single value of $x_{1}$. You have to multiply by the probability of that value of $x_{1}$ and sum over all potential choices for $x_{1}$. – Mr. F Mar 20 '12 at 0:56
Nevermind, I see what you did – lord12 Mar 20 '12 at 1:00
You're right. The mistake though is not quite what you mentioned. It is that in my first conditional expectation, I forgot to divide by the marginal probability $P(X_{1} = x_{1})$. Then, multiplying by that again will cancel it to make the last formula correct. I am updating to reflect this. – Mr. F Mar 20 '12 at 1:06

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