Question

The joint probability mass function of (X, Y) is given by p(x, y) = k(2x + 3y),

x = 0, 1, 2; y = 1, 2, 3. Find all the marginal and conditional probability

distributions. Also find the probability distribution of (X + Y)

Answer 1

$$\sum_x\sum_yp(x,y)=1$$ $$\begin{matrix} & & & y \\ & p(x, y) & 1 & 2 & 3 \\ \hdashline & 0 & 3k & 6k & 9k \\ x & 1 & 5k & 8k & 11k \\ & 2 & 7k & 10k & 13k \\ \end{matrix}$$

$$3k+6k+9k+5k+8k+11k+7k+10k+13k=1$$

$$k=\dfrac{1}{72}$$

$$x=0: p_X(0)=3k+6k+9k=18k=\dfrac{1}{4}$$

$$x=1: p_X(1)=5k+8k+11k=24k=\dfrac{1}{3}$$

$$x=2: p_X(2)=7k+10k+13k=30k=\dfrac{5}{12}$$

$$p_X(x)=\begin{cases} 1/4 &\text{if } x=0 \\ 1/3 &\text{if } x=1 \\ 5/12 &\text{if } x=2 \\ 0 &\text{otherwise } \\ \end{cases}$$

$$y=1: p_Y(1)=3k+5k+7k=15k=\dfrac{5}{24}$$

$$y=2: p_Y(2)=6k+8k+10k=24k=\dfrac{1}{3}$$

$$y=3: p_Y(3)=9k+11k+13k=33k=\dfrac{11}{24}$$

$$p_Y(y)=\begin{cases} 5/24 &\text{if } y=1 \\ 1/3 &\text{if } y=2 \\ 11/24 &\text{if } y=3 \\ 0 &\text{otherwise } \\ \end{cases}$$

$$p_{X|Y}(x|y)=\dfrac{p(x, y)}{p_X(x)}$$

Conditional pmf's of $X$given $Y$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c} x & p_{X|Y}(x|1) & p_{X|Y}(x|2) & p_{X|Y}(x|3) \\ \hline 0 & 1/6 & 1/3 & 1/2 \\ \hdashline 1 & 5/24 & 1/3 & 11/24 \\ \hdashline 2 & 7/30 & 1/3 & 13/30 \\ \hdashline \end{array}$$

$$p_{Y|X}(yx)=\dfrac{p(x, y)}{p_Y(y)}$$

Conditional pmf's of $Y$given $X$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c} y & p_{Y|X}(y|0) & p_{Y|X}(y|1) & p_{Y|X}(y|2) \\ \hline 1 & 1/5 & 1/3 & 7/15 \\ \hdashline 2 & 1/4 & 1/3 & 5/12 \\ \hdashline 3 & 3/11 & 1/3 & 13/33 \\ \hdashline \end{array}$$

$$x+y=1, (0,1):p(1)=3k=\dfrac{1}{24}$$

$$x+y=2, (0,2), (1, 1):p(2)=6k+5k=\dfrac{11}{72}$$

$$x+y=3, (0,3),(1,2),(2,1):p(3)=9k+8k+7k=\dfrac{1}{3}$$

$$x+y=4, (1,3),(2,2):p(4)=11k+10k=\dfrac{7}{24}$$

$$x+y=5, (2,3):p(15)=13k=\dfrac{13}{72}$$

$$\def\arraystretch{1.5} \begin{array}{c:c} x+y & 1 & 2 & 3 & 4 & 5 \\ \hline p(x+y) & 1/24 & 11/72 & 1/3 & 7/24 & 13/72 \end{array}$$