How to find the conditional expectation for this pdf 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 $E[X_2|X_1=x_1]$
Attempt:
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.
 A: 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.
