Download 21-325 (Fall 2008): Homework 6 (TWO sides) Due by

21-325 (Fall 2008): Homework 6 (TWO sides)
Due by October 29, in class
Show FULL JUSTIFICATION for all your answers.
1. Let X be a positive continuous random variable having density fX . Find a formula for the
density of Y = 1/(1 + X).
Solution. To compute the density of Y , we first compute the c.d.f. of Y , then we get the p.d.f.
of Y by taking the derivative of the c.d.f. of Y . Let y ∈ (0, 1), since Y can only take values in
(0, 1).
P (Y ≤ y) = P (1/(1 + X) ≤ y) = P (X ≥ 1/y − 1) = 1 − P (X ≤ 1/y − 1).
Then by talking the derivative and the chain rule, we get fY (y) = fX (1/y − 1)y −2 .
2. Suppose that a continuous random variable X satisfies P (X > a + b) = P (X > a)P (X > b) for
all a ≥ 0, b ≥ 0.
(a) What can you say about the distribution of X ?
Solution. P (X > a + b) = P (X > a)P (X > b) implies the memoryless property of X. Then X
has an exponential distribution, since it is the only distribution.
(b) What can you say about the distribution of bXc, the greatest integer less than or equal to
X ?
Solution. From (a), we know X has an exponential distribution. Since bXc can only take
integer values, then bXc is a discrete random variable. It’s p.m.f. is given by
pbXc (k) = P (k ≤ X < k + 1) =
Z k+1
λe−λx dx = e−λk − e−λ(k+1) for k = 0, 1, ....
k
3. Let X be a continuous random variable with p.d.f.
f (x) = 4x3 if 0 < x < 1 and f (x) = 0 otherwise.
Let Y = 1 − 3X 2 .
(a) Calculate the expectation of Y using the formula for the expectation of a function of a
continuous random variable X with density fX :
Z ∞
E[g(X)] =
−∞
1
g(x)fX (x) dx.
Solution.
Z 1
E[Y ] =
(1 − 3x2 )4x3 dx = −1.
0
(b) Find the p.d.f. of Y = 1 −
3X 2 .
Solution. Since X can only take values in (0, 1), then 0 < X 2 < 1, and then 0 < 3X 2 < 3, and
then −3 < 3X 2 < 0, and then −2 < 1 − 3X 2 < 1. So Y takes values in (−2, 1). Let y ∈ (−2, 1).
P (Y ≤ y) = P (X 2 ≥ (1 − y)/3) = P (X ≥
q
q
(1 − y)/3) + P (X ≤ − (1 − y)/3),
p
where P (X ≤ − (1 − y)/3) = 0 since X can only take values in (0, 1). By differentiating on both
sides, we get
q
q
fY (y) = −fX ( (1 − y)/3)( (1 − y)/3)0 = (2 − 2y)/9.
(c) Calculate the expectation of Y using the definition of expectation and the answer in (b).
Solution.
Z 1
E[Y ] =
y(2 − 2y)/9dy = −1.
−2
4. Let X be a normal random variable with mean 0 and variance 1 and let Y be a normal
random variable with mean 1 and variance 4. (For questions (a) and (c), you may use the table for
the standard normal distribution on the web at http://www.sjsu.edu/faculty/gerstman/EpiInfo/ztable.htm).
(a) Find P (X ≤ 1.5) and P (X ≤ −1).
Solution. P (X ≤ 1.5) = 0.9332 and P (X ≤ −1) = P (X ≥ 1) = 1 − P (X ≤ 1) = 1 − 0.8413.
(b) Find the p.d.f of (Y − 1)/2.
Solution. By the normalization, (Y −1)/2 is a normal random variable with mean 0 and variance
1. Then it density is
2
√1 e−x /2 .
2π
(c) Find P (−1 ≤ Y ≤ 1).
Solution. −1 ≤ Y ≤ 1 implies that −1 < (Y − 1)/2 < 0. Then P (−1 ≤ Y ≤ 1) = P (−1 <
(Y − 1)/2 < 0) = 0.5 − (1 − 0.8413) = 0.3413.
5. Show that for a non-negative continuous random variable X, for n ∈ N ,
E[X n ] =
Z ∞
nxn−1 P (X > x)dx.
0
2
Solution. X n =
RX
0
RX
nxn−1 dx. Then E[X n ] = E[X n ] = E[
0
R∞
nxn−1 dx] = E[
0
1{x≤X} nxn−1 dx].
By exchanging expectation and integration,
E[X n ] =
Z ∞
0
E[1{x≤X} ]nxn−1 dx =
Z ∞
nxn−1 P (X > x)dx.
0
6. Consider two continuous random variables Y and Z, and a random variable X that is equal to
Y with probability p and to Z with probability 1 − P .
(a) Show that the p.d.f. of X is given by
fX (x) = pfY (x) + (1 − p)fZ (x).
Solution. P (X ≤ x) = pP (Y ≤ x) + (1 − p)P (Z ≤ x). Then by differentiating both sides, we
get fX (x) = pfY (x) + (1 − p)fZ (x).
(b) Calculate the c.d.f. of the two-sided exponential random variable that has p.d.f. given by
fX (x) = pλeλx if x < 0 and f (x) = (1 − p)λe−λx if x ≥ 0.
Solution. When x ≤ 0, FX (x) = peλx . When x > 0, FX (x) =
R0
∞ pλe
λy dy +
Rx
0
(1 − p)λe−λy dy =
p + (1 − p)(1 − e−λx ).
7. For a > 0, let X be a uniformly distributed random variable on [0, a]. Find the cumulative
distribution function and expected value of Y = min(X, a/2).
Solution. Since X takes values in [0, a], then Y takes values in [0, a/2]. If y ≤ 0, FY (y) = 0. If
0 < y < a/2, FY (y) = P (Y ≤ y) = P (X < y) = y/a. If y ≥ a/2, FY (y) = 1.
Z a
E[Y ] =
Z a/2
min(x, a/2)1/a dx =
0
= 3a/8.
0
8. If Y is uniformly distributed over [0, 5], what is the probability that the roots of the equation
4x2 + 4xY + Y + 2 = 0 are both real?
Solution. We need (4Y )2 − 4 ∗ 4 ∗ (Y + 2) ≥ 0. It follows that 16Y 2 − 16Y − 32 ≥ 0, and then
(Y + 1)(Y − 2) ≥ 0. Then we need to compute P ((Y + 1)(Y − 2) ≥ 0). Then we need to compute
P (Y ≤ −1 or Y ≥ 2). Since Y ≥ 0, then we need to compute P (Y ≥ 2), which is 3/5. So the
probability is 3/5.
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