NTUEE Probability and Statistics - Part 3. Homework Solution
Problem 1.
Suppose you roll two four-faced dice, with faces labeled 1, 2, 3, 4, and each equally likely to appear on top. Let X denote the smaller of the two numbers that appear on top. (If both dice show the same number, then X is equal to that common number.) Please find the probability mass function (PMF) of X.
Sol.
X = 4 corresponds to: (4, 4).
X = 3 corresponds to: (3, 3), (3, 4), (4, 3).
X = 2 corresponds to: (2, 2), (2, 3), (2, 4), (3, 2), (4, 2).
X = 1 corresponds to: (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (3, 1), (4, 1).
So pX(4) = 1/16, pX(3) = 3/16, pX(2) = 5/16, pX(1) = 7/16, and pX = 0 otherwise.
Problem 2.
According to the PMF in Problem 1, find P[X < 4|X > 1].
Sol.
P[X < 4|X > 1] = P[1 < X < 4]
P[X > 1] = 8/16 9/16 = 8
9.
Problem 3.
According to the PMF in Problem 1, find E(2X).
Sol.
E(2X) = 21· 7
16+ 22· 5
16+ 23· 3
16+ 24· 1
16 = 14 + 20 + 24 + 16
16 = 74
16 = 37 8 . Problem 4.
Given that E[X + 4] = 10 and E[(X + 4)2] = 116, what is the standard deviation of X?
Sol. E[X + 4] = E[X] + 4 = 10 ⇒ E[X] = 6,
E[X2+ 8X + 16] = E[X2] + 8E[X] + 16 = 116 ⇒ E[X2] = 52, σX =pVar(X) = pE[X2] − E2[X] =√
16 = 4.
Problem 5.
Let X be a continuous random variable with PDF f (x) = 4xe−2x for x > 0 and
1
f (x) = 0 otherwise. Please find the mean and the variance of X.
Sol. The identityR∞
0 tne−tdt = n! follows from integration by parts. By the substi- tution t = 2x,
E(X) = Z ∞
0
4x2e−2xdx = 1 2
Z ∞ 0
t2e−tdt = 2!
2 = 1.
Similarly,
E(X2) = Z ∞
0
4x3e−2xdx = 1 4
Z ∞ 0
t3e−tdt = 3!
4 = 3/2.
So Var(X) = E(X2) − E2(X) = 3/2 − 1 = 1/2.
Problem 6.
Let Y = X4, where X follows an exponential distribution with mean equal to λ1. Find the probability density function of Y .
Sol.
Since X is exponential distributed,
fX(x) = λeλx x ≥ 0, 0 otherwise.
For t ≥ 0, P[X4 ≤ t] =Rt1/4
0 λeλxdx = 1 − e−λt1/4. We have FY(y) = 0
P[X4 ≤ y] = 0 y < 0, 1 − e−λy1/4 y ≥ 0.
Taking derivative with respect to y, fY(y) =
λ
4y−3/4e−λy1/4 y ≥ 0,
0 y < 0.
Problem 7.
Let U be a uniform random variable on [0, 1], and let V = U1. (a) Find the density function of V . (b) What is the mean of V ?
Sol.
First we find the CDF,
FV(v) = P(V ≤ v) = P (1/U ≤ v) = P (U ≥ 1/v) = Z 1
1/v
dv = 1 − 1/v
2
for v ≥ 1. Thus, fV(v) = Fv0(v) = 1/v2 for v ≥ 1 and fV(v) = 0 otherwise. We have E(V ) =
Z ∞
−∞
vfV(v)dv = Z ∞
1
v
v2dv = ln v|∞1 = ∞, so the mean does not exist.
Problem 8.
Given a random variable X with CDF FX and let g be the CDF of X, i.e. g(x) = FX(x). Suppose that g is strictly increasing. We can define another random variable Z by Z = g(X). Please find the CDF of Z.
Sol.
For 0 ≤ z ≤ 1, we have
FZ(z) = P[Z ≤ z] = P[FX(X) ≤ z]
= P[X ≤ FX−1(z)]
= FX(FX−1(z))
= z since FX is strictly increasing.
Thus,
FZ(z) =
0 if z < 0, z if 0 ≤ z ≤ 1, 1 if z > 1.
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