5. (10 points) Random variables X and Y have following distributions (o) (1 poin

Question

5. (10 points) Random variables X and Y have following distributions (o) (1 point) Find the moment generating fiunction of the following random variable. P(X =-2) = 1/2, P(X = 3) = 1/3,P(X = 5) = 1 /6 (b) (1 point) Find the moment generating function of the following random variable. P(Y =-1)-1/2, P(Y = 2-1/4, P(Y = 4-1/4 (c) (1 point) Use the moment generating function to find E|X] (d) (1 point) Use the moment generating function to find E[X] (e) (1 point) Use the moment generating function to find E[X3] (f) (1 point) Use the moment generating function to find ElY] (g) (1 point) Use the moment generating function to find EY

Explanation / Answer

a) The MGF for X here is obtained as:

Mx(t) = E[ ext ] = 0.5e-2t + (1/3)e3t + (1/6)e5t

This is the required MGF for X here.

b) The MGF for Y here is obtained as:

My(t) = E[ eyt ] = 0.5e-t + 0.25e2t + 0.25e4t

This is the required MGF for Y here.

c) Differentiating the MGF for X, we get:

Mx'(t) = -1e-2t + e3t + (5/6)e5t

Now the expected value of X here is computed as:

E(X) = Mx'(0) = -1 + 1 + (5/6) = (5/6)

Therefore E(X) = 5/6

d) The above derivative of MGF is again differentiated here to get:

Mx''(t) = 2e-2t + 3e3t + (25/6)e5t

The second moment of X here is obtained as:

E(X2) = Mx''(0) = 2 + 3 + (25/6) = 9.1667

e) Again the MGF is differentiated and 0 is inserted to get the third moment of X as:

E(X3) = Mx'''(0) = -4 + 9 + (125/6)= 25.8333

f) In a similar way to X, the three moments of Y are also obtained here as:

E(Y) = My'(0) = -0.5 + 0.5 + 1 = 1

g) E(Y2) = My''(0) = 0.5 + 1 + 4 = 5.5

h) E(Y3) = My'''(0) = -0.5 + 2 + 16 = 17.5

i) Now the expected value of X + Y here is computed as:

E(X + Y) = E(X) + E(Y) = (5/6) + 1 = 11/6 = 1.8333

Therefore E(X + Y) = 1.8333

Related Questions

Navigate

• Browse (All)
• Browse 5
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.