Conditional expectation and expected value Given the joint distribution $f(x,y) =  {x \choose y} (\frac{1}{2})^x \frac{x}{20}$ where $y=0,1,2 \cdots x$ and $x=2,3,4,5,6$.
Find $E(y\mid x)$ and $E(y)$
Wondering if I am missing something.
Using Binomial identity and direct definition of $f(x)$ (the limit was from the plot I made):
$f(x) =  (\frac{1}{2})^x \frac{x}{20}  \sum_{y=0}^x   {x \choose y}  =(\frac{1}{2})^x \frac{x}{20}  2^{x} = \frac{x}{20} $
Use $f(x)$ to compute $f(y\mid x)$, given that the joint distribution is given:
So that $f(y\mid x) = \frac{f(x,y)} {f(x)} = \frac{ {x \choose y} (\frac{1}{2})^x \frac{x}{20}}{ \frac{x}{20}} = {x \choose y} 2^{-x} $
$E(y \mid x) = \sum_{y=0}^{x} y f(y\mid x) = 2^{-x}  \sum_{y=0}^{x} {x \choose y} y   $
Then: Using the identity : $  \sum_{y=0}^{x} {x \choose y} y  = x 2^{x-1}  $
$E(y \mid x) = \sum_{y=0}^{x} y f(y\mid x) = 2^{-x}  \sum_{y=0}^{x} {x \choose y} y  = x (2^{-x} \cdot 2^{x-1}) = \frac{x}{2} $
Does this makes sense?
How does $E(y)$ follows, I guess $E(y) = E(E(y \mid x))$
$E(y) = E(E(y \mid x))$
$E(y) = E(E(y \mid x)) = \sum_{x=2}^6 \frac{x}{20} (2^{-x} \sum_{y=0}^{x} {x \choose y} y) =  \frac{1}{40}\sum_{x=2}^6 x^2  $

The easiest  way to proceed is to observe that your joint pmf is
$$p(x,y)=p(y|x)p(x)=\binom{x}{y}\left(\frac{1}{2}\right)^x\times\frac{x}{20}$$
Thus $(Y|X=x)\sim Bin\left(x;\frac{1}{2}\right)$
And X is a discrete rv with support $x=\{2,3,4,5,6\}$ and pmf
$$p(X=x)=x/20$$
Thus
$$\mathbb{E}[Y|X=x]=x/2$$
And
$$\mathbb{E}[Y]=\mathbb{E}[\mathbb{E}[Y|X]]= \mathbb{E}[X/2]= \frac{1}{2 }\mathbb{E}[X]
 = \frac{1}{2 }\times\frac{9}{2}=\frac{9}{4}$$
