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Department of EECS - University of California at Berkeley
EECS 126 - Probability and Random Processes - Spring 2007
Midterm 2: 4/03/
SOLUTIONS
The random variables X, Y are independent. X is uniformly distributed in [0, 1] and Y is ex-
ponentially distributed with mean 1, so that P (Y > y) = e
−y for y ≥ 0. Let Z = min{X, Y }.
Calculate the mean and the variance of Z.
Hint: You may need the following intermediate results. Let a n
∫ 1
0
x
n e
−x dx for n ≥ 0. One
finds that a 0 = 1 − e
− 1 , a 1 = 1 − 2 e
− 1 , a 2 = 2 − 5 e
− 1 , a 3 = 6 − 16 e
− 1 .
We have P (Z > x) = P (X > x, Y > x) = P (X > x)P (Y > x) = (1 − x)e
−x for x ∈ [0, 1]. The
density fZ of Z is thus given by
fZ (x) = −
d
dx
P (Z > x) = e
−x
−x
= 2e
−x
− xe
−x
, for 0 ≤ x ≤ 1.
Now,
E(Z) =
∫ 1
0
xf Z
(x)dx = 2
∫ 1
0
xe
−x dx−
∫ 1
0
x
2 e
−x dx = 2a 1
−a 2
= 2− 4 e
− 1 −[2− 5 e
− 1 ] = e
− 1 ≈ 0. 37
and
E(Z
2 ) =
∫ 1
0
x
2 f Z
(x)dx = 2
∫ 1
0
x
2 e
−x dx −
∫ 1
0
x
3 e
−x dx = 2a 2
− a 3
= 2[2 − 5 e
− 1 ] − [6 − 16 e
− 1 ]
= −2 + 6e
− 1
,
so that
var(Z) = −2 + 6e
− 1
− e
− 2
≈ 0. 07.
The random variables X and Y are as in Problem 1.
a) Calculate f V
(v) where V = X + Y.
b) Calculate E(V ) and var(V ).
a) To find the density of V we note that, if v < 1, then
P (V > v) = P (X > v)+
∫ v
0
fX (x)P (Y > v−x)dx = 1−v+
∫ v
0
e
x−v
dx = 1−v+(1−e
−v
) = 2−v−e
−v
.
Also, if v > 1,
P (V > v) =
∫ 1
0
f X
(x)P (Y > v − x)dx =
∫ 1
0
e
x−v
dx = e
−v
[e − 1].
We find the density of V by differentiating and we get
f V
(v) =
{
1 − e
−v , for v ≤ 1
e
−v (e − 1), for v > 1.
b) We find
E(V ) = E(X) + E(Y ) =
and
var(V ) = var(X) + var(Y ) = E(X
2 ) − E(X)
2
2 ) − E(Y
2 ) =
since
E(Y
2
) =
∫ ∞
0
y
2
e
−y
dy = −
∫ ∞
0
y
2
de
−y
=
∫ ∞
0
2 ye
−y
dy = 2.
The random variables X, Z 1
, Z
2
,... are independent; X is N (0, σ
2 ) and Z n
= N (0, u
2 ) for n ≥ 1.
Let
Xn = E[X|Y 1 ,... , Yn] where Yk = X + Zk, k = 1,... , n. You must choose the number n of
measurements so that P (|X −
X
n
a) Find
X
n
b) Use Chebyshev’s inequality to estimate the smallest value of n you should use if σ
2 = 4, u
2 = 1
so that P (|X −
X
n
c) Use the Gaussian distribution to estimate the smallest value of n you should use if σ
2 = 4, u
2
1 so that P (|X −
X
n
Note: Here are some potentially useful values: P (|N (0, 1)| > 1 .64) = 10%, P (|N (0, 1)| > 1 .96) =
5%, P (|N (0, 1)| > 2 .58) = 1%.
a) We know, by symmetry, that
Xn = a(Y 1 + · · · + Yn).
Also, a should be such that X −
X
n
is orthogonal to each Y k
. Writing that X −
X
n
⊥ Y
1
we find
0 = E((X − a(nX + Z 1
+ · · · + Z
n
))(X + Z
1
)) = (1 − an)σ
2 − au
2 ,
so that
a =
σ
2
u
2
2
b) We find that
E((X −
X)
2 ) = var((1 − an)X − aZ 1
− · · · − aZ n
) = (1 − an)
2 σ
2
2 u
2
u
4 σ
2
(u
2
2 )
2
nu
2 σ
4
(u
2
2 )
2
u
2 σ
2
u
2
2
Using Chebyshev’s inequality, we have
P (|X −
Xn| > ) ≤
E((X −
X)
2 )
2
u
2 σ
2
2 (u
2
2 )
With σ
2 = 4, u
2 = 1, = 0.1, we see that we need
0 .01(1 + 4n)
which implies n ≥ 2 , 000.
c) We know that X −
X
2 = N (0, b
2
n
) where b
2
n
u
2 σ
2
u
2 +nσ
2
. Now,
P (|N (0, b
2
n
)| > ) = P (|N (0, 1)| >
b n
) ≤ 0 .05 if
b n
Hence, we need /b n
≥ 1 .96, or b n
≤ / 1 .96, or b
2
n
2 /(1.96)
2 ≈
2 /4. That is,
1 + 4n
so that n ≥ 400. The example shows that the Chebyshev estimate is a bit pessimistic.
Assume that Z 1
, Z
2
are independent, zero mean, and with respective variances 1, 4. The random
variable X is independent of {Z 1 , Z 2 } and is equally likely to be equal to −1 or +1.
a) Find
X = M AP [X|X + Z 1 , X + Z 2 ].
b) Find P (
X 6 = X).
Hint: Here are some values of F W
(x) for W = N (0, 1) that you may need:
x -1.6 -1.5 -1.4 -1.3 -1.2 -1.1 -1.0 -0.9 -0.8 -0.7 -0.
F
W
(x) 0.055 0.067 0.081 0.097 0.115 0.136 0.159 0.184 0.212 0.242 0.
a) Let Y 1 = X + Z 1 and Y 2 = X + Z 2. We find
f [Y|X]
[y|1] =
2 π
e
−(y 1
−1)
2 / 2
2 π 4
e
−(y 2
−1)
2 / 8
and
f [Y|X]
[y| − 1] =
2 π
e
−(y 1 +1)
2 / 2
2 π 4
e
−(y 2 +1)
2 / 8
.
Hence,
L(y) =
f [Y|X]
[y|1]
f [Y|X]
[y| − 1]
= exp{ 2 y 1
y 2
Accordingly,
M AP [X|Y] =
{
1 , if 2Y 1
1
2
Y
2
− 1 , otherwise.
b) By symmetry,
P (
X 6 = X) = P [
X = +1|X = −1] = P [2Y
1
Y
2
> 0 |X = −1].
Now, if X = −1, 2Y 1
1
2
Y
2
= N (− 2. 5 , 4 +
1
4
× 4 = 5). Also,
P (N (− 2. 5 , 5) > 0) = P (N (0, 1) >
) = P (N (0, 1) > 1 .12) ≈ 0. 13 ,
according to the values given in the table.
