I need help with E through I 81. * Consider the function g(x, y) given by, 1 1.5

Question

I need help with E through I

81. * Consider the function g(x, y) given by, 1 1.5 2.53 11/4 0 0 0 2 01/8 0 0 y 30 1/4 00 400 1/4 0 5 0 0 0 1/8 and complete / determine the following: (a) Show that g(x,y) satisfies the properties of a joint pmf. (See Table in §6.01) (b) P(X 4.7) (t) Ex], EY], Var(X) (g) Marginal probability distribution of X (h) Conditional probability distribution of Y given that X -1.5 (i) Conditional probability distribution of X given that Y -2 (j) EY | X = 1.5] (k) Are X and Y independent? , Var(Y)

Explanation / Answer

e) The required probability here is computed as:

P( X > 1.8, Y > 4.7 ) = P(X = 3, Y = 5) = 1/8 = 0.125

Therefore 0.125 is the required probability here.

f) The marginal distribution for X here is computed as:

P(X = 1) = 0.25, P(X = 1.5) = (1/8) + (1/4) = 0.375, P(X = 2.5) = 0.25 and P(X = 3) = 0.125

Therefore, we get:

E(X) = 0.25*1 + 0.375*1.5 + 0.25*2.5 + 0.125*3 = 1.8125

Therefore, we get: E(X) = 1.8125

The second moment of X here is computed as:

E(X) = 0.25*12 + 0.375*1.52 + 0.25*2.52 + 0.125*32 = 3.78125
Therefore variance here is computed as: Var(X) = E(X2) - (E(X))2 = 3.78125 - 1.81252 = 0.4961

Therefore we get: var(X) = 0.4961

Now the marginal distribution for Y here is obtained as:

P(Y = 1) = 0.25,
P(Y = 2) = 0.125
P(Y = 3) = 0.25
P(Y = 4) = 0.25
P(Y = 5) = 0.125

Therefore, we have here:

E(Y) = 0.25 + 2*0.125 + 3*0.25 + 4*0.25 + 5*0.125 = 2.875

Therefore, we get: E(Y) = 2.875

The second moment here is computed as:

E(Y) = 0.25 + 22*0.125 + 32*0.25 + 42*0.25 + 52*0.125 = 12.125

Therefore the variance here is computed as:

Var(Y) = E(Y2) - (E(Y))2 = 12.125 - 2.8752 = 3.8594

Therefore, we get: Var(Y) = 3.8594

g) The marginal distribution for X here is already computed in previous part as:

P(X = 1) = 0.25, P(X = 1.5) = (1/8) + (1/4) = 0.375, P(X = 2.5) = 0.25 and P(X = 3) = 0.125

h) Given X = 1.5, the conditional probability distribution for Y here is obtained as:

P(X = 1.5) = 0.375

Now the conditional probability of Y here is computed as:

P(Y = 1 | X =1.5) = 0
P(Y = 2 | X =1.5) = 0.125 /0.375 = 0.3333
P(Y = 3 | X =1.5) = 0.25 /0.375 = 0.6667
P(Y = 4 | X =1.5) = 0
P(Y = 5 | X =1.5) = 0

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