Department of Electrical Engineering and Computer Sciences
EECS 126: Probability and Random Processes
Problem Set 12
Spring 2018
Self-Graded Scores Due: 5 PM, Friday May 4, 2018
Submit your self-graded scores via th
...
Department of Electrical Engineering and Computer Sciences
EECS 126: Probability and Random Processes
Problem Set 12
Spring 2018
Self-Graded Scores Due: 5 PM, Friday May 4, 2018
Submit your self-graded scores via the Google form:
https://goo.gl/forms/HYw8QwfrHNKf0czx2.
Make sure you use your SORTABLE NAME on CalCentral.
1. Geometric MMSE
Let N be a geometric random variable with parameter 1 − p, and (Xi)i∈N be
i.i.d. exponential random variables with parameter λ. Let T = X1 + · · · + XN .
Compute the LLSE and MMSE of N given T.
Solution:
First, we calculate P(N = n | T = t), for t > 0 and n ∈ Z+.
P(N = n | T = t) =
P(N = n)fT|N (t | n)
P∞
k=1 P(N = k)fT|N (t | k)
=
(1 − p)p
n−1λ
n
t
n−1
e
−λt/(n − 1)!
P∞
k=1(1 − p)p
k−1λkt
k−1e−λt/(k − 1)!
=
λ(λpt)
n−1/(n − 1)!
λ
P∞
k=1(λpt)
k−1/(k − 1)! =
(λpt)
n−1
e
λpt(n − 1)!, n ∈ Z+.
Next, we calculate E[N | T = t].
E[N | T = t] = X∞
n=1
n
(λpt)
n−1
e
λpt(n − 1)!
=
X∞
n=1
(λpt)
n−1
e
λpt(n − 1)! +
X∞
n=1
(n − 1) (λpt)
n−1
e
λpt(n − 1)!
= 1 +
λpt
e
λpt X∞
n=2
(λpt)
n−2
(n − 2)! = 1 +
λpt
e
λpt e
λpt = 1 + λpt.
Hence, the MMSE is E[N | T] = 1 + λpT. The MMSE is linear, so it is also
the LLSE.
In terms of a Poisson process, T represents the first arrival of a marked Poisson
process with rate λ, where arrivals are marked independently with probability
1 − p. The marked Poisson process has rate λ(1 − p). The unmarked points
form a Poisson process of rate λp. In time T, the expected number of unmarked
points is λpT, so the conditional expectation of the number of points at time
T, N, is 1 + λpT.
1
2. Property of MMSE
Let X, Y1, . . . , Yn be square integrable random variables. Argue that
E
(X − E[X | Y1, . . . , Yn])2
≤ E
X −
Xn
i=1
E[X | Yi
]
2
.
Solution:
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