Download Mathematics 426 Robert Gross Homework 7 Answers 1. Suppose

Mathematics 426
Robert Gross
Homework 7
Answers
1. Suppose that X is a Poisson random variable with parameter λ. Show that E[Xn ] =
λE[(X + 1)n−1 ]. Use this result to compute E[X], E[X2 ], and E[X3 ].
Answer : We have
E[Xn ] =
∞
X
kn P(X = k) =
k=0
∞
X
=λ
∞
X
λk X n −λ λk X n−1 −λ λk
=
k e
=
k
e
k!
k!
(k − 1)!
k=1
k=1
∞
kn e−λ
k=0
∞
∞
X
λ
λk
=λ
(k + 1)n−1 e−λ
= λE[(X + 1)n−1 ]
(k − 1)!
k!
k=0
k−1
k
n−1 −λ
e
k=1
Now we have E[X] = λE[10 ] = λ, and then E[X2 ] = λE[(X + 1)1 ] = λE[X + 1] = λ(λ + 1) =
λ2 + λ, and nally E[X3 = λE[(X + 1)2 ] = λE[X2 + 2X + 1] = λ((λ2 + λ) + 2λ + 1) =
λ(λ2 + 3λ + 1) = λ3 + 3λ2 + λ.
2. (a ) Write the power series expansion for eλ + e−λ .
(b ) Suppose that X is a Poisson random variable with parameter λ. Show that
P(X is even) = (1 + e−2λ )/2.
Answer : (a ) We have
λ
e +e
=
∞
X
λk
k=0
(b ) We have
P(X is even) =
−λ
∞
X
∞
X
λ2k
+
(−1)
=2
k! k=0
k!
(2k)!
k=0
P(X = 2k) = e
k=0
∞
X
−λ
kλ
k
λ
∞
X
e + e−λ
1 + e−2λ
λ2k
−λ
=e
=
.
(2k)!
2
2
k=0
3. Whenever Alice and Bob have lunch, they decide who will pay for the meal by each
rolling a die. The one who rolls the lower number must pay. If there is a tie, they roll
again, and continue rolling until one rolls a lower number than the other.
What is the expected number of rolls needed to determine who will pay for lunch?
Answer : The probability of a tie is 16 , so the probability that the game ends in a single roll
is 56 . Because this problem describes a geometric distribution, we know that the expected
value is p1 = 56 .
4. Suppose that X is a continuous random variable with probability density function
f(x) =
c(1 − x4 )
0
−1 < x < 1
otherwise
(a ) What is c?
(b ) What is the cumulative distribution function for X?
Answer : (a ) We compute
Z1
x5
(1 − x ) dx = x −
5
−1
4
1
−1
8
= .
5
Therefore, c = .
(b ) If we denote the cumulative distribution function as F(a), we know that F(a) = 0
for a < −1 and F(a) = 1 for a > 1. For −1 6 a 6 1, we have
5
8
5
F(a) =
8
Za
5
(1 − x ) dx =
8
−1
4
x5
x−
5
a −1
5
=
8
a5 4
a−
+
5
5
5. Suppose that X is a binomial random variable with E[X] = 6 and Var(X) = 2.4. Compute
P(X = 5).
Answer : We have np = 6 and np(1 −p) = 2.4. Division yields 1 − p = 264 = 0.4, p = 0.6,
and n = p6 = 10. Now, P(X = 5) = 10
0.65 0.45 ≈ 0.2007.
5
.
6. A lling station is provided with gasoline once each week. Suppose that the weekly
volume of sales in thousands of gallons is a random variable with probability density
function
f(x) =
6(1 − x)5
0<x<1
otherwise
What must the capacity of the lling station's tank be so that the probability of the station
running out of gas in a given week is 0.01?
Answer : One way to think about this problem is in terms of the cumulative distribution
function F(a): we wish to nd a so that F(a) = 0.99. We know that for a 6 0 6 1, we
have
Za
a
F(a) =
6(1 − x)5 dx = −(1 − x)6 0 = 1 − (1 − a)6 .
0
0
We then must solve (1 − a)6 = 0.01, or 1 − a ≈ 0.4642, or a ≈ 0.5358. This is the capacity
of the station's tank, in thousands of gallons.
7. Remember that a poker hand contains 5 random cards. Let X be the number of suits
in a hand. Find P(X = k) for k = 1, 2, 3, and 4. Hint : See Ch. 2, self-help prob. 11, for a
discussion of P(X = 4).
Answer : The simplest to compute is P(X = 1): pick a suit, and then the cards in that
suit.
4 · 13
33
P(X = 1) = 525 =
≈ 0.0020.
5
16660
For X = 2, we can either use Inclusion{Exclusion, or else rely on the fact that a two-suited
hand must have either 4-1 distribution or else 3-2 distribution. For Inclusion{Exclusion,
we compute the probability that the hand contains no more than 2 suits, and then subtract
the one-suited hands:
26
5
52
5
4
·
P(X = 2) =
2
13
5
52
5
4
·
−2
2
=
143
≈ 0.1459.
980
Alternatively, we can compute the probability of the two dierent distributions mentioned
above and add:
13 13
13 13
4·3
P(X = 2) =
+
2
3
1
52
5
4
=
143
980
Inclusion{Exclusion is trickier when X = 3 (see if you can work it out on your own), but
fortunately, there are again only 2 possible distributions: 2-2-1 and 3-1-1:
P(X = 3) =
13
2
6·2
13
2
13
+
1
52
5
13
3
13
1
13
1
=
4901
≈ 0.5884.
8330
Finally, for X = 4, the simplest approach is one that the text does not oer. The only
possible distribution is 2-1-1-1:
P(X = 4) =
4
13
2
13 13
1
1
52
5
13
1
=
2197
≈ 0.2637.
8330
The 4 probabilities sum to 1, as they must.
8. Let Y be a random variable. Show that
Z∞
E[Y] =
Z∞
P(Y > y) dy −
P(Y < −y) dy
0
by showing that
0
Z∞
Z∞
P(Y > y) dy =
0
Z∞
xfY (x) dx
0
Z0
xfY (x) dx.
P(Y < −y) dy = −
0
−∞
In these formul, fY (x) is the probability density function for Y .
Answer : I prefer to work in the other direction, assuming that the probability density
function for Y is f(y) and starting with the formula for E[Y]:
Z∞
E[Y] =
Z0
Z∞
xf(x) dx =
−∞
Z∞ Zx
=
xf(x) dx +
−∞
0
xf(x) dx =
0
Z0 Z0
dy f(x) dx −
0
Z∞
Z0
xf(x) dx −
0
(−x)f(x) dx
−∞
dy f(x) dx
−∞ x
In the rst integral, we have 0 6 y 6 x 6 ∞, while in the second integral we have
−∞ < x < y < 0. Changing the order of integration:
Z∞ Z∞
=
Z0 Zy
f(x) dx dy −
0
y
Z∞
f(x) dx dy =
−∞ −∞
Z0
P(Y > y) dy −
0
P(Y < y) dy
−∞
One more step: let y = −z in the second integral, with dy = −dz:
Z∞
=
Z0
P(Y > y) dy +
Z∞
P(Y < −z) dz =
∞
0
Z∞
P(Y > y) dy −
0
P(Y < −z) dz.
0
9. If X is a random variable with density function f(x), show that
Z∞
g(x)f(x) dx.
E[g(X)] =
−∞
Hint : The previous exercise Zshows that
Z∞
∞
E[g(X)] =
P(g(X) > y) dy −
0
P(g(X) < −y) dy.
0
Now proceed as we did in class.
Answer : Write f(x) as the probability density function for X. We have
Z∞
E[g(X)] =
Z0∞
=
0
Z∞
P(g(X) > y) dy −
P(g(X) < −y) dy
0
Z
Z∞ Z
f(x) dx dy −
f(x) dx dy
x:g(x)>y
0
x:g(x)<−y
Again, we switch the order of integration in each of the integrals, noting in the second
integral that if g(x) < −y, then −g(x) > y:
Z
Z g(x)
Z
Z −g(x)
dy f(x) dx −
dy f(x) dx
x:g(x)<0 0
Z
Z
=
g(x)f(x) dx −
(−g(x))f(x) dx
x:g(x)>0
x:g(x)<0
Z
Z
=
g(x)f(x) dx +
g(x)f(x) dx
x:g(x)>0
x:g(x)<0
Z
=
g(x)f(x) dx
=
x:g(x)>0 0
x:g(x)6=0
Finally, notice that
Z
g(x)f(x) dx = 0, because the integrand is 0:
x:g(x)=0
Z
Z
=
Z∞
g(x)f(x) dx +
x:g(x)6=0
g(x)f(x) dx =
x:g(x)=0
g(x)f(x) dx
−∞