Solutions - 5190-02

(a)

Find $x$ using the constraint ${\sum }_{x,y}{P}_{X,Y}(x,y)=1$:

$$\begin{array}{c}0.1+5(0.05)+0.15+2(0.2)+x=1\\ \\ \gg \gg x=1-0.1-0.25-0.15-0.4\gg \gg x=0.1\end{array}$$

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(b)

Find the marginal PMF of $X$ by summing rows:

$$P_X(x)=\{\begin{array}{cc}0.1+0.15+0.05=0.3& x=1\\ 2(0.2)+2(0.05)=0.5& x=2\\ 2(0.05)+0.1=0.2& x=3\end{array}$$

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(c)

(1) Define the possible values of $Z$:

Since $X\in \{1,2,3\}$ and $Y\in \{0,1,2,3\}$, we have that $Z\in \{0,1,2,3,4,6,9\}$.

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(2) Define PMF of $Z$:

Go through each possible value of $Z$ and see when it occurs.

$$P_Z(z)=\{\begin{array}{cc}P[Y=0]& z=0\\ P[X=1\cap Y=1]& z=1\\ P[X=1\cap Y=2]+P[X=2\cap Y=1]& z=2\\ P[X=1\cap Y=3]+P[X=3\cap Y=1]& z=3\\ P[X=2\cap Y=2]& z=4\\ P[X=2\cap Y=3]+P[X=3\cap Y=2]& z=6\\ P[X=3\cap Y=3]& z=9\end{array}$$

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(3) Substitute values for each probability:

$$P_Z(z)=\{\begin{array}{cc}0.35& z=0\\ 0.15& z=1\\ 0.05& z=2\\ 0.05& z=3\\ 0.05& z=4\\ 0.3& z=6\\ 0.05& z=9\end{array}$$

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(d)

(1) Find $P[X=Y]$:

$$\begin{array}{c}P[X=1\cap Y=1]+P[X=2\cap Y=2]+P[X=3\cap Y=3]\\ \\ \gg \gg 0.15+0.05+0.05\gg \gg 0.25\end{array}$$

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(2) Find $P[X>Y]$:

$$\begin{array}{c}P[Y=0]+P[X=2\cap Y=1]+P[X=3\cap Y=1]+P[X=3\cap Y=2]\\ \\ \gg \gg 0.35+0.05+0.05+0.05\gg \gg 0.5\end{array}$$