UC Berkeley
Department of Electrical Engineering and Computer Science
EE 126: Probablity and Random Processes
Problem Set 5
Spring 2007
Issued: Thursday, March 8, 2007 Due: Thursday, March 15, 2007
Problem 5.1
1. N = 200, 000.
2. N = 100, 000.
Problem 5.2
a)This is a straight application of the Chebychev inequality. Chebychev tells us that:
P (|X − 7| ≥ 3) ≤ V3ar 2 1
2 = 3 and therefore that P (4 < X < 10) ≥ 3 .
V ar
b)If the variance = 9, P (|X − 7| ≥ 3) ≤ 32 = 1 and therefore that P (4 < X < 10) ≥ 0.
Problem 5.3
(a)
1/4k if k = 1, 2, 3, 4 and n = 1, . . . , k
pN,K (n, k) =
0 otherwise
(b)
1/4 + 1/8 + 1/12 + 1/16 = 25/48 n=1
1/8 + 1/12 + 1/16 = 13/48 n=2
pN (n) = 1/12 + 1/16 = 7/48 n=3
1/16 = 3/48 n=4
0 otherwise
(c) The conditional PMF
6/13
k=2
pN,K (2, k) 4/13 k=3
pK|N (k|2) = =
pN (2)
3/13 k=4
0 otherwise
(d) Let A be the event that Chuck bought at least 2 but no more than 3 books, E[K|A] = 3
var(K|A) = 35
21
(e) E[T ] = 4
1
Problem 5.4
1. 1
20 , x=1
3
20 , x=2
3 13
pX (x) = 10 , x=3 , E[X] =
1
4
, x=4
2
0 , otherwise
2.
1
10 , w=2
1
, w=3
20
7
20 , w=4
pW (w) = 3
10 , w=6
1
, w=8
5
0 , otherwise
3. E[R] = 34 , var(R) = 63
80
√
3
4. σR|A = 4
Problem 5.5
1. µ = 3, σ 2 = 2
2. µ = 5, σ 2 = 20.
Problem 5.6
1. To calculate the probability of heads on the next flip, we use the continuous version of
the total probability theorem:
Z 1
P (H) = P (H|P = p) · fP (p)dp
0
Z 1
= p2 ep dp
0
= e−2
2
2. Here we need to use the continuous version of Bayes’ theorem:
P (A|p)fP (p)
fP |A (p|A) = R1
P (A|p)fP (p)dp
(0
p2 ep
e−2 , 0≤p≤1
=
0, otherwise
3. Now we apply the above result to the technique used in (a):
Z 1
P (H) = P (A|P = p) · fP |A (p|A)dp
0
Z 1
1
= p3 ep dp
e−2 0
1
= · (6 − 2e)
e−2
.564
=
.718
≈ .786
Problem 5.7
Note that we can rewrite E[X1 | Sn = sn , Sn+1 = sn+1 , . . .] as follows:
E[X1 | Sn = sn , Sn+1 = sn+1 , . . .]
= E[X1 | Sn = sn , Xn+1 = sn+1 − sn , Xn+2 = sn+2 − sn+1 , . . .]
= E[X1 | Sn = sn ],
where the last equality holds due to the fact that the Xi ’s are independent.
We also note that
E[X1 + · · · + Xn | Sn = sn ] = E[Sn | Sn = sn ] = sn
It follows from the linearity of expectations that
E[X1 + · · · + Xn | Sn = sn ] = E[X1 | Sn = sn ] + · · · + E[Xn | Sn = sn ]
Because the Xi ’s are identically distributed, we have the following relationship:
E[Xi | Sn = sn ] = E[Xj | Sn = sn ], for any 1 ≤ i ≤ n, 1 ≤ j ≤ n.
Therefore,
sn
E[X1 + · · · + Xn | Sn = sn ] = nE[X1 | Sn = sn ] = sn ⇒ E[X1 | Sn = sn ] = .
n
3
Problem 5.8
a) γ is determined because the density function must integrate to 1. Since (X, Y ) uni-
formly distributed in R, we have: ZZ
γdxdy = 1
where the integral is over the area of R. Therefore γ1 =Area R.
b) Showing independence is equivalent to showing that:
fX,Y (x, y) = fX (x) · fY (y)
But this is clear, since:
fX,Y (x, y) = fY (x) = fY (y) = 1
c) Consider a region that consists of the upper half plane, and the point (1, −1). If we are
told that Y < 0, X is determined, and hence X, Y cannot be independent in this region.
d) To find the probability that (X, Y ) lie in the circle C inscribed in the R in part (b) we
could integrate, or observe that the integral will in fact come out to the area of the circle,
and hence the desired probability will be the ratio of the are of the circle to the area of the
square:
.25π
P ((X, Y ) ∈ C) = .
1
