Download Math 151 Midterm 2 Solutions

Math 151 Midterm 2 Solutions
1.
Given a random variable X with mean µ and variance σ 2 > 0. Construct a random variable f (X)
with mean 1 and variance 2.
Solution.
We will use two identities:
E[aX + b] = aE[X] + b for any real numbers a and b
V ar(cX + d) = c2 V ar(X) for any real numbers c and d
First, we
will transfrom the variance. √Apply the above identities:
√
Then σ2 X has variance 2 and mean σ2 µ.
Applying√them again,
mean changes and we get:
√
f (X) = σ2 X − σ2 µ + 1 has mean 1 and variance 2.
2.
P∞
For a nonnegative integer-valued random variable N , show that i=0 iP [N > i] = 12 (E[N 2 ] − E[N ])
Solution.
P∞ i2 −i
1
2
i=0 2 P [N = i]
2 (E[N ] − E[N ]) =
P∞
Now, let’s count number of P[N=i]-s for each i in the expression Pi=0 iP [N > i]
∞
This is exactly sum of js for each 1 6 j < i, because P [N > j] = i=j+1 P [N = i]
2
= i 2−i .
Hence numbers of P [N = i]-s is 1 + 2 + ... + i − 1 = (i−1)i
2
Since this is for each i, the two expressions are equal.
3.
X and Y are independent exponential random variables each having parameter λ. Let Z = X + Y ,
V = eX be random variables. Find the joint density function of Z and V .
Solution.
g1 (x, y) = x + y, g2 (x, y) = ex
Partial derivatives of g1 with respect to x and y are both 1 and partial derivatives of g2 with respect
to x and y are ex and 0, respectively. Then |J| = ex . Page 300 of textbook gives:
fZ,V (z, v) = fX,Y (x, y)|J|−1 = λ2 e−λ(x+y) e−x = λ2 e−λz v −1 when lnv > 0 and z − lnv > 0,
0, otherwise.
Notice that final restrictions are direct consequence of the fact that expontial r.v is nonzero for
x = lnv > 0 and y = z − x = z − lnv > 0
4.
−y
The joint density of X and Y is given by f (x, y) = e y , 0 < x < y, 0 < y < ∞
Compute E[X 2 |Y = y].
Solution.
Ry
R y −y
fY (y) = 0 f (x, y)dx = 0 e y = e−y
f (x,y)
f (y)
Ry
y] = 0
fX|Y (x|y) =
E[X 2 |Y =
= y1 , 0 < x < y, 0 < y < ∞
Ry 2
x2 fX|Y (x|y)dx = 0 xy dx =
1 x3 y
y 3 |x=0
=
y2
3
5.
An airline company sells 200 tickets for a plane with 198 seats, knowing that the probability a
passenger will not show up for the flight is 0.01. Use the Poisson approximation to compute the
probability they will have enough seats for all passengers who show up.
Solution.
In the language of Poisson approximation we have n = 200, p = 0.01
Hence, λ = np = 2
If N is the number of passenger who do not show up. We want:
P (N > 2) = 1 − P [N = 0] − P [N = 1] = 1 − e−2 1 − e−2 2 = 1 − 3e−2
2
6.
The time required to repair a machine is an exponential random variable with λ = 1 Find:
(i) The probability that the repair time exceeds 3 hours.
(ii) Given that its duration exceeds 8 hours, what is the conditional probability that a repair time
takes at least 10 hours.
Solution.
R∞
(i) P [X > 3] = 3 e−x dx = e−3 T
(X>8)]
e−10
−2
(ii) P [X > 10|X > 8] = P [(X>10)
= PP[X>10]
P [X>8]
[X>8] = e−8 = e