Math 419 – Spring 2016
Expectation
Problem Set 9
Expectation (Rules to know)
1. Double Expectation Theorem : For any two random variables X and Y.
E(Y) = E(E(Y|X))
2. Similar result for computing Variance:
Var(Y) = E(Var(Y|X)) + Var(E(Y|X))
Note: The two results above are particularly useful when X and Y are quite different
distributions (e.g X is discrete and Y is continuous)
3. Covariance : of the random variables X and Y is a measure of their tendency to vary together.
Cov(X,Y) = E [(X – E(X)) · (Y – E(Y))]
= E(XY) – E(X) · E(Y)
4. a) Var(X) = Cov(X,X)
b) Cov(X,Y) = Cov(Y,X)
c) Cov(aX+bY, Z) = aCov(X,Z) + bCov(Y,Z)
d) Cov(X,Y) = 0, if X and Y are independent
e) Var(aX+bY) = Cov(aX+bY, aX+bY)
= a2Var(X) + b2Var(Y) + 2abCov(X,Y)
5. Correlation Coefficient: of random variables X and Y is given by r(rho)
r = Cov(X,Y) / (sx · sy)
6. Properties of r:
a) –1 £ r £ 1
; r = 0 if X and Y are independent (but converse is not always true)
b) If r = 1, X and Y are perfectly positively correlated.
c) If r = – 1, X and Y are perfectly negatively correlated
d) If r = 0, there is no discernible LINEAR relationship between X and Y.
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Math 419 – Spring 2016
Expectation
Problem Set 9
1. The number of traffic accidents has a distribution N which is a Poisson random variable with mean
L, where L is uniformly distributed over [0, 3]. Find Var( N ).
2. Suppose the number of claims in a year is Poisson with mean 900 and that the average claim amount
is 1000 with a standard deviation of 500. Calculate the expected value and the standard deviation of
the aggregate claim for one year.
3. Suppose the number of claims in a year is Poisson with mean 900 and that the average claim amount
is 1000 with a standard deviation of 500. Calculate the expected value and the standard deviation of
the aggregate claim for one year.
4. The remaining lifetimes of husband and wife are modeled by f(x, y) = 1/600, where 0 < x < 30, and
0 < y < 20. Find the expected time until both die.
5. Suppose the joint probability density function of X and Y is f(x, y) = (4x)/y, for 0 < x < y < 1 and
zero otherwise. Find Var(Y | X = ¼).
6. A random variable X has the cumulative distribution function
Calculate the expected value of X.
7. There are two persons in a car. The probability of each of them being injured in a crash is 0.3. If
injured, the losses of each are uniformly distributed on [0, 1]. What is the expected number of
persons injured, given that the total loss is less than 1?
8. Let X denote the size of a bodily injury claim and Y denote the size of the corresponding property
damage claim. Let Z1 = X + Y. We are given that Var(X) = 144, Var(Y) = 64 and Var(X + Y) = 308.
It is expected that bodily injury claims will rise 10% next year and property damage will rise by a
fixed amount of 5. Let Z2 be the new trial of bodily injury and property damage. Compute
Cov(Z1, Z2 ).
9. Let X1, X2, X3 be uniform random variables on the interval (0, 1) with
Cov(Xi, Xj) = 1/24 for i, j = 1, 2, 3 and i ¹ j. Calculate the variance of X1 + 2X2 – X3.
10. A couple buys two 10-year term life insurance policies with a death benefit of 25,000 on each
policy. Both the wife and the husband have a 90% chance of surviving 10 years independent of
each other. Find the variance for the total death benefit paid on the two policies combined.
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Math 419 – Spring 2016
Expectation
Problem Set 9
11. X and Y are modeled inflation rates, in terms of %, for two different countries at the end
of a five-year period. X is uniformly distributed on the interval (0, 10). Y, given X = x, is
uniformly distributed on the interval (0, x).
Calculate Cov(X, Y).
12. The losses for a certain product are given by Z = 2X + 5Y + 1, where X and Y are random variables
with Var(X) = 4, Var(Y) = 6, and Var(X + Y) = 14. What is the variance of Z?
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