Probability Theory, Math 170b, Winter 2013, Tonci Antunovic - Homework 3

From the textbook solve the problems 29, 30, 31, 32 and 33 from the Chapter 4.Solve the problems 1, 2, 4, 5 and 6 from the Chapter 4 additional exercises at

http://www.athenasc.com/prob-supp.html

And also the problems below:

Problem 1. Given a positive random variable X assume that the transform of the randomvariable Y = logX is given by

MY (s) =es

4− 2es+

1

4+e−3s

4.

Use the definition of the transform of Y to compute E[√X].

Solution: We have

E[√X] = E

[e

12logX

]= E

[e

12Y]

= MY (1/2) =e1/2

4− 2e1/2+

1

4+e−3/2

4.

Problem 2. Consider the function f(s) = (sin s+ 2)2. Compute the 2nd derivative f ′′(s) andby analyzing the sign of f ′′(s) explain why f(s) can’t be a transform MX(s) of some randomvariable X.

Solution: The second derivative is f ′′(s) = 2 cos2 s − 2 sin2 s − 4 sin s = 2 cos(2s) − 4 sin swhich is negative for say s = π/2. However, the 2nd derivative of a transform MX(s) can’t benegative since

MX(s) = E[X2esX

],

and since the expression under the expectation X2esX is never negative.

Problem 3. Assume that the transform of a continuous random variable X satisfies MX(s) =MX(−s). Show that its PDF fX satisfies fX(t) = fX(−t). (Hint: compare the transforms andPDFs of X and −X).

Solution: Let’s find the transform of −X:

M−X(s) = E[es(−X)

]= E

[e−sX

]= MX(−s).

By the assumption this is the same as MX(s) so M−X(s) = MX(s). Therefore, random variablesX and −X have the same distribution and since they are both continuous they have the samePDF so f−X(t) = fX(t). However, it always holds that f−X(t) = fX(−t). This is because

F−X(t) = P(−X ≤ t) = P(X ≥ −t) = 1− FX(−t) ⇒ f−X(t) = fX(−t),

where the implication follows by differentiation with respect to t. Therefore, fX(−t) = f−X(t) =fX(−t).

1

Problem 4. You are raising funds for a good cause and so you ask n people to give you money.Each person is equally likely to give you either 0 or 1 or 2 dollars independently of all otherpeople. If X is total amount of money you got, compute E[X], var(X) and E[X3].

Solution: If Xi represents the amount of the money the ith person gave you then E[Xi] = 1and var(Xi) = 2/3 the total amount of money X = X1 + · · ·+Xn satisfies

E[X] = n, var(X) = 2n/3.

For the 3rd moment we don’t have simple formulas, so let’s try to use the transform. Thetransform of Xi is MXi(s) = 1

3(1 + es + e2s) so MX(s) = 13n (1 + es + e2s)n. The 3rd derivative

comes after some computation

M(3)X (s) =

n(n− 1)(n− 2)

3n(1 + es + e2s)n−3(es + 2e2s)3

+ 3n(n− 1)

3n(1 + es + e2s)n−2(es + 2e2s)(es + 4e2s) +

n

3n(1 + es + e2s)n−1(es + 8e2s).

At s = 0 the value is

M(3)X (0) = n(n− 1)(n− 2) + 5n(n− 1) + 3n = n3 + 2n2.

Therefore E[X3] = n3 + 2n2.

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