Question

Problem 2. Consider the following joint probabilities for the two variables X and Y. 1 2 3 .14 .25 .01 2 33 .10 .07 3 .03 .05 .02 Find the marginal probability distribution of Y and graph it. Show your calculations. b. Find the conditional probability distribution of Y (given that X = 2) and graph it. Show your calculations. c. Do your results in (a) and (b) satisfy the probability distribution requirements? Explain clearly. d. Find the correlation coefficient between the Y and X. Show your calculations. e. Comment on the strength and direction of association between Y and X. f. Find both the expected value and the standard deviation of (X+Y). Show your calculations.

Answer 1

$a)f_X(x)=\left\{\begin{matrix} 0.5 & x=1\\ 0.4 & x=2\\ 0.1 & x=3 \end{matrix}\right.$

(0.4 y = 1 fr(y) = {0.5 y=2 10.1 y = 3

b)P(Y|X = 2) = = 0.625 Y = 1 = 0.25 Y => 0.05 = 0.125 Y = 3

c)Yes, the results satisfy probability distribution requirements because the sum of the probabilities=1

d)E[X]=1*0.5+2*0.4+3*0.1=1.6

E[Y]=1*0.4+2*0.5+3*0.1=1.7

$Corr=\frac{\sum \sum f_{X,Y}(x,y)(X-1.6)(Y-1.7)}{\sqrt{\sum f_X(x)(X-1.6)^2}\sqrt{\sum f_Y(y)(Y-1.7)^2}}=0$

e)The correlation of x,y=0 which means X and Y are independent.

f)Lets first calculate the probability distribution of X+Y

$f(X+Y)=\left\{\begin{matrix} 0.14 &X+Y=2 \\ 0.58 &X+Y=3 \\ 0.14 &X+Y=4 \\ 0.12 &X+Y=5 \\ 0.02 &X+Y=6 \end{matrix}\right.$

E[X+Y]=2*0.14+3*0.58+4*0.12+5*0.12+6*0.02=3.3

SD(X+Y) = f(X+Y)*(X+Y-3.3)2 = V0.85 = 0.922