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University of Illinois, Urbana–Champaign Fall 2007 ECE 313: Solutions to Homework 3
- N is a discrete random variable.
E[N ] =
∑^ ∞
n=
n Pr(N = n)
= Pr(N = 1) + Pr(N = 2) + Pr(N = 2) + · · · + Pr(N = n) + · · · + Pr(N = n) ︸ ︷︷ ︸ n
∑^ ∞
n=
Pr(N = n) +
∑^ ∞
n=
Pr(N = n) + · · · +
∑^ ∞
n=i
Pr(N = n) + · · ·
= Pr(N ≥ 1) + Pr(N ≥ 2) + · · · + Pr(N ≥ i) + · · ·
∑^ ∞
n=
Pr(N ≥ n)
∑^ ∞
n=
n Pr(N > n) = Pr(N = 2) + Pr(N = 3) + Pr(N = 4) + · · · + Pr(N = n) + · · ·
- 2 Pr(N = 3) + 2 Pr(N = 4) + · · · + 2 Pr(N = n) + · · ·
- 3 Pr(N = 4) + · · · + 3 Pr(N = n) + · · · · · ·
- (n − 2) Pr(N = n − 1) + (n − 2) Pr(N = n) + · · ·
- (n − 1) Pr(N = n) + · · ·
= Pr(N = 2) + 3 Pr(N = 3) + · · · +
n(n − 1) 2
Pr(N = n) + · · ·
∑^ ∞
n=
(n^2 Pr(N = n) − n Pr(N = n)) + Pr(N = 1) − Pr(N = 1)
∑^ ∞
n=
n^2 Pr(N = n) −
∑^ ∞
n=
n Pr(N = n))
Using LOTUS, we have E[N 2 ] =
n=1 n
(^2) Pr(N = n), then
(E[N 2 ] − E[N ])
- (a) Let pY be the pmf of Y. Since g is a one-to-one function, for any x there exists a unique y = g(x) such that pX (x) = pY (y). This is true because the same probability measure is defined on the function projection of X. Let the random variable Y = g(X) be the projection space of X via function g. We have,
E[g(X)] = E[Y ] =
y
ypY (y)
g(x)
g(x)pY (g(x))
Since pY (g(x)) = pX (x) for all x, we have
x
g(x)pX (x)
(b) If g is a many-to-one function, for all x 1 , x 2 , · · · , xn such that y = g(x 1 ) = g(x 2 ) = · · · = g(xn), we have pY (y) =
xi pX^ (xi). The probability mass for these^ y’s that have many to one mappings is the sum of the mass of all inverse imagines. Therefore,
E[g(X)] = E[Y ] =
y
ypY (y)
y
xi,y=g(xi)
g(xi)pX (xi )
x
g(x)pX (x)
- Using LOTUS (the result from 16), we get
E[Y ] = E[
X − μ σ
] =
E[X] − μ σ
We know that var(X) = E[X^2 ] − (E[X])^2. So E[X^2 ] = σ^2 + μ^2. By using this, we have
E[Y 2 ] = E[
σ^2
(X^2 − 2 Xμ + μ^2 )] =
σ^2
(E[X^2 ] − 2 E[X]μ + μ^2 ) = 1
- (a) At each round, let R be the event that red shows. The sample space for a game of maximum three tries is shown in the first row of the following table. For each outcome, we compute net earning and the probability. Ω R Rc, R, R Rc, Rc, Rc^ Rc, Rc, R Rc, R, Rc X $1 $1 -$3 -$1 -$ Pr[X] (^1838203818381838203820382038203820381838203818382038) For the winning probability, we sum the outcomes that X > 0. Pr(X > 0) = 1838 + 20 38
18 38
18 38 = 0.5918. (b) You have more than 50% chance to win the game. However the answer from (c) shows that you lose 11 cents on average when you play the game. (c) E[X] = 1838 + 203818381838 − 3 382020382038 − 203820381838 − 203818382038 = − 36139 = − 0. 1080
- (a) Show that 0 ≤ pX (k) ≤ 1. We noticed that pX (k) is a monotonously increasing function with k and pX (k) is always greater than 0. When k = n, pX (n) = (^) n+1^2. As n ≥ 1, the largest value pX (n) ≤ 1. Then we need to show that the probability mass is summed up to 1.
∑n k=1 pX^ (k) =^
2 n(n+1)
n(n+1) 2 = 1. Therefore,^ pX^ is a valid pmf. (b) The range of Y is all the inverse of integers from [1, n]. The pmf of Y is
pY (k) = Pr(Y = k) = Pr(1/X = k) = Pr(X = 1/k) =
kn(n + 1)
with k = 1/i where 1 ≤ i ≤ n is an integer. (c) E[Y ] =
k kpY^ (k) =^
2 n+1. Or by LOTUS,^ E[Y^ ] =^
k
1 k pX^ (k) leads to the same answer.
