Let $P(X=1)=P(X=2)=P(Y=1)=P(Y=2)=\frac{1}{2}$. Find the smallest and largest possible value of $E[XY]$.
My solution:
If $X$, $Y$ are independent, then $E[XY] = (\frac{3}{2})^2 = \frac{9}{4}$.
Now, let's assume they are dependent in such a way:
- If $X=1$ then $Y=1$ and if $X=2$ then $Y=2$, then $E[XY] = \frac{1}{2}\cdot 1 \cdot 1 + \frac{1}{2}\cdot 2 \cdot 2 = \frac{5}{2}$.
- If $X=1$ then $Y=2$ and if $X=2$ then $Y=1$, then $E[XY] = \frac{1}{2}\cdot 1 \cdot 2 + \frac{1}{2}\cdot 2 \cdot 1 = 2$.
Hence, $2 \leq E[XY] \leq \frac{5}{2}$
Is this a correct answer?
You have only shown that $E[XY]$ can attain the values $2$ and $2.5$, but you have not shown that $E[XY]$ is always in the interval $[2,2.5]$. Here is one way to complete the proof.
Let $p=P(X=1\wedge Y=1)$. Since $$ 1/2=P(X=1)=P(X=1\wedge Y=1)+P(X=1\wedge Y=2), $$ we conclude that $P(X=1\wedge Y=2)=1/2-p$. Similarly, you can show $$ P(X=2\wedge Y=1)=1/2-p,\qquad P(X=2\wedge Y=2)=p. $$ Therefore, $$ \begin{align} E[XY] &=p\cdot (1\cdot1)+(1/2-p)\cdot (1\cdot 2)+(1/2-p)\cdot (2\cdot 1)+p\cdot (2\cdot 2) \\&=p+2(1-2p)+4p=2 + p \end{align}$$ Since $0\le p\le 1/2$, we conclude that $2\le E[XY] \le 2+1/2$, as desired.