STA 131A Second Midterm (Fall 2018)
November 30, 2018
STUDENT NAME:
STUDENT ID:
Problem 1 [20 pts]
Let the r.v. X denote the number of successes in 24 independent binomial trials with
probability p of success, and let Y denote the number of failures (so that X + Y = 24).
(i) What is the distribution of X?
(ii) Write out an expression for the probability P (X = 3Y ).
(iii) Write out an expression for the probability P (X > 3Y ) in a way that the appro-
priate Tables can be used.
(iv) Compute the probability P (X = 4Y ).
Problem 2 [22 pts]
The r.v. X has the p.d.f. f (x) = λe−λx for x > 0 and f (x) = 0 for x ≤ 0, λ > 0.
(i) Compute the d.f., F (x), of the r.v. X, as well as the 1 − F (x).
(ii) Give the expression for the probability P (1 ≤ X ≤ 8) in terms of λ.
(iii) Find a simple expression for the conditional probability P (X > 10|X > 3) in
terms of λ.
Hint: For parts (ii) and (iii), use part (i).
Problem 3 [18 pts]
Let the r.v. X be distributed as N (µ, σ2) and define the r.v. Y by: Y
= cX + d,
where c and d are constants.
It is given that the m.g.f. of X, MX , is given by:
(
σ2t2 )
MX = exp
µt +
, t∈R
2
(i) Find the m.g.f. of Y , MY , by using familiar properties of an m.g.f..
(ii) Conclude that Y ∼ N (cµ + d, (cσ)2), and justify your arguments.
Problem 4 [22 pts]
The p.d.f. of the r.v. Y is given by: fY (y) = 1 for 0 < y < 1, and 0 otherwise, whereas
the conditional p.d.f. of X, given Y
= y, is given by: fX|Y (x|y) =1
y for0<x<y,
and 0 otherwise.
(i) Find the joint p.d.f. of X and Y , fX,Y (x, y), and specify the domain of x and y.
(ii) Find the (marginal) p.d.f. of X, fX (x), and specify the domain of x.
(iii) Find the conditional expectation E(X|Y = y), and specify the domain of y.
(iv) Use part (iii) (and a familiar property of a conditional expectation) in order to
find EX.
Problem 5 [18 pts]
Let X and Y be r.v.’s such that:
EX = 1, EX2 = 3, EY = 2, EY2 = 29, E(XY ) = −4.
(i) Compute the σ2(X) and the σ2(Y ).
(ii) Compute the covariance Cov(X, Y ) and the correlation coefficient ρ(X, Y ).
(iii) Compute the covariance Cov(X, 2X + Y − 3).
