Let $(X,Y)$ be a bivariate discrete R.V. with joint pmf $p(x,y)=1/{{m+1 \choose 2}}$

Question

Let $(X,Y)$ be a bivariate discrete R.V. with joint pmf $$p(x,y)=\begin{cases} \frac{1}{{m+1 \choose 2}} \text{if $y=1,2 \ldots,x$ & $x=1,2,\ldots,m$} \\ 0 \ \text{otherwise} \end{cases}$$ for a given positive integer $(m>1)$.Find $E(X)$ from the marginal distribution of $X$ and also from $E(X|Y)$.

I got the first part .Let the marginal pmf of $X$ be $p_X(x)$. Thus, $p_X(x)=\sum_{y} \frac{1}{{m+1 \choose 2}}=\frac{x}{{m+1 \choose 2}}$ Thus $E(X)=\sum_{x=1}^{m} \frac{x^2}{{m+1 \choose 2}} = \frac{2m+1}{3}$

But how to get $E(X)$ from $E(X|Y)$?

Siong Thye Goh · Accepted Answer · 2017-12-02T00:32:17.020

1

We can use the total law of expectation: $$E(X)=\sum_y E(X|Y=y)P_Y(y)$$

If $y \in \{ 1, \ldots, m\}$,

\begin{align}P_Y(y)&=\sum_{x=y}^m p(x,y) =(m-y+1) \frac{1}{\binom{m+1}{2}}\\ \end{align}

$X|Y\sim \operatorname{Uni}\{Y, m\},$

Hence $$E(X|Y=y)=\frac{y+m}{2}$$

\begin{align}E(X)&=\sum_{y=1}^m \frac{y+m}{2}\cdot\frac{m-y+1}{\binom{m+1}{2}} \\ &=\frac{1}{m(m+1)}\sum_{y=1}^m(m^2-y^2+y+m)\\ &= \frac{1}{m(m+1)}\left[(m^2+m)m-\sum_{y=1}^my^2+\sum_{y=1}^my\right]\\ &=\frac{1}{m(m+1)}\left[m^2(m+1)-\frac{m(m+1)(2m+1)}{6} +\frac{m(m+1)}{2}\right]\\ &=m-\frac{2m+1}{6}+\frac12\\ &=\frac{4m+2}{6}\\ &=\frac{2m+1}{3}\end{align}

edited Dec 02 '17 at 00:32

answered Nov 04 '17 at 16:26

Siong Thye Goh

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Can you find me $E(X|Y)$? And is my evaluation $E(X)$ correct? – Nov 04 '17 at 16:36
Your $E(X)$ seems fine. to find $E(X|Y)$, think of the following, given a particular $Y$ value, what are the possible values that $X$ can take. – Siong Thye Goh Nov 04 '17 at 16:40
If I calculate $p_{X|Y}(x|y)=\frac{1}{m}$. Then can I calculate $E(X|Y)=\sum_{x=1}^{m} x.p_{X|Y}(x|y)$? – Nov 04 '17 at 16:47
I would be pleased if you work it out for me .. I am a little puzzled – Nov 04 '17 at 16:53
let say $y=2$, $x$ does take value $1$ with positive probability. – Siong Thye Goh Nov 04 '17 at 16:53
Yes, but how will it help me calculate E(Y|X)? – Nov 04 '17 at 16:58
Given a particular $Y$ value, what are the possible values that $X$ can take. $X|Y$ follows a discrete uniform distribution. You just have to find the possible ranges and you can compute the expectation. – Siong Thye Goh Nov 04 '17 at 17:04
Let us continue this discussion in chat. – Nov 05 '17 at 03:01
Anyone please post a complete solution – Dec 01 '17 at 16:35

Let $(X,Y)$ be a bivariate discrete R.V. with joint pmf $p(x,y)=1/{{m+1 \choose 2}}$

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