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- Homework Statement
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- Relevant Equations
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The probability mass function of X and Y:

$$p_{X,Y}(1,1)= \frac{1}{2}\cdot \frac{3}{4}=\frac{3}{8}$$

$$p_{X,Y}(1,0)=\frac{1}{2}\cdot \frac{1}{4}=\frac{1}{8}$$

$$p_{X,Y}(0,1)= \frac{1}{2}\cdot \frac{3}{4}=\frac{3}{8}$$

$$p_{X,Y}(0,0)= \frac{1}{2}\cdot \frac{1}{4}=\frac{1}{8}$$

or

$$

p_{X,Y}(a,b)=

\begin{cases}

\frac{3}{8} & (a,b)=(1,1) \quad \text{or}\quad(a,b)=(0,1) \\

\frac{1}{8} & (a,b)=(1,0) \quad \text{or}\quad(a,b)=(0,0)

\end{cases}$$

(b)

The marginal distribution of X is

$$p_X(1)=p_{X,Y}(1,0)+p_{X,Y}(1,1)=\frac{1}{2}\frac{3}{4}+\frac{1}{2}\frac{1}{4}=\frac{1}{2}$$

$$p_X(0)=p_{X,Y}(0,0)+p_{X,Y}(0,1)=\frac{1}{2}\frac{1}{4}+\frac{1}{2}\frac{3}{4}=\frac{1}{2}$$

The marginal distribution of Y is

$$p_Y(1)=p_{X,Y}(0,1)+p_{X,Y}(1,1)=\frac{1}{2}\frac{3}{4}+\frac{1}{2}\frac{1}{4}=\frac{3}{4}$$

$$p_Y(0)=p_{X,Y}(0,0)+p_{X,Y}(1,0)=\frac{1}{2}\frac{1}{4}+\frac{1}{2}\frac{3}{4}=\frac{1}{4}$$

(c)

The expected value of Y is

$$E(Y)=1\cdot p_Y(1)+0\cdot p_Y(0)=\frac{3}{4}$$

(d)

$$E((X-0.5)\times Y)=\sum\limits_{(a,b)}(a-.5)bp_{X,Y}(a,b)$$

$$(1-0.5)(1)p_{X,Y}(1,1)+(1-0.5)(0)p_{X,Y}(1,0)+$$

$$(0-0.5)(1)p_{X,Y}(0,1)+(0-0.5)(0)p_{X,Y}(0,0)=$$

$$\frac{1}{2}\cdot\frac{3}{8}+0-\frac{1}{2}\cdot \frac{3}{8}+0=0$$