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UNIVERSITY OF ILLINOIS, URBANA-CHAMPAIGN
Department of Electrical and Computer Engineering ECE 313: Probability with Engineering Applications-Fall 2003 Problem Set 4 Solution
Problems to be turned in:
- Requires independence! Independence is unlikely in practice: if one chip is defective, it seems likely that there is a bad batch.
- Let (a, b, c) be the numbers appearing on the 3 dice, 1 ≤ a, b, c ≤ 6. The number of outcomes for which a + b + c = i is equal to the number of ways in which i identical balls can be divided into 3 partitions. This can be done in
(i− 1 2
ways, by choosing 2 of the (i − 1) spaces between the balls, and inserting separators there. Therefore,
P (X = i) =
(i− 2 1 ) 216 ,^ i^ ≥^3 0 , i < 3 , (impossible events).
Note that this approach is invalid for i ≥ 9. For instance, for i = 9, (7, 1 , 1) is a valid partition, but is not a permissible combination of the die values.
- Let g(x) = Ixi (x) and h(y) = Iyi (y). Then,
P (X = xi) · P (Y = yi) = E[g(X)] · E[h(Y )] by definition of expectation = E[g(X)h(Y )] by assumption = E[Ixi (X)Iyi (Y )] by definition of g and h = P (X = xi and Y = yi) by definition of expectation.
- If X is the amount that you win, then P (X = 1.1) = 49 = 1 − P (X = −1).
(a) E[X] = (1.1) 49 − 59 = − 09.^6 ≈ − 0 .067. (b) V ar(X) = (1.1)^2 · 49 + 59 − ( 09.^6 )^2 ≈ 1 .089. (c) Let S = X 1 +X 2 +· · ·+X 10. Then, E[S] = 10E[Xi] = − 2 /3 and V ar(S) = 10V ar(Xi) = 10.89. Chebyshev’s bound gives,
P (S ≥ 3) = P (S + 0. 67 ≥ 3 .67) ≤ P (|S + 0. 67 | ≥ 3 .67)
≤ V ar(S)
- 672
=^10.^89
In this example, we obtain a tighter upper bound using Markov’s bound. Note that Xi ≥ − 1 for each i, so S ≥ −10. Consequently, by Markov’s inequality,
P (S ≥ 3) = P (S + 10 ≥ 13)
≤
E[S + 10]
13 =^
13 = 0.^72.
- (a) The number of wrong answers can be modeled as a binomial random variable W with param- eters (N − K, 56 ).
(b) Since P (W = n) =
(N −K
n
( 56 )n( 16 )N^ −K−n, for n = 0, 1 ,... , N − K, for fixed N and n, P (W = n) is a function of K. This is the likelihood function, denoted `(K) in lecture. Consideration of the ratio as follows provides a formula for the maximum:
(K)(K − 1)
(N −K
n
( 56 )n( 16 )N^ −K−n (N −(K−1) n
( 56 )n( 16 )N^ −(K−1)−n
= (N^ −^ K)(N^ −^ K^ −^ 1)^ · · ·^ (N^ −^ K^ −^ n^ + 1) (N − (K − 1))(N − (K − 1) − 1) · · · (N − (K − 1) − n + 1) (^16)
if and only if N^ N− −KK−+1n+1 ≥ 16 , that is, K ≤ N − 65 n + 1. Since K is an integer, the ML estimator can be expressed as K̂ = bN − 65 n + 1c, where b·c is a floor function. (c) Perhaps. It will be interesting to see what the current ECE 313 students think about this grading policy!
- (a) I(X > c) = exp{α(X − c)}, ∀α > 0 , c ∈ R. E[I(X > c)] ≤ E[exp{α(X − c)}] = E[eαX^ ]e−αc.
(b) X =
i=1 Bi, where^ Bi^ = 0 or 1 with probability^
1
E[eαX^ ] = E
[
exp
α
∑^10
i=
Bi
)]
∏^10
i=
E[exp(αBi)]
E[eαBi^ ] =^1 2
eα·^0 +^1 2
eα·^1
=^1 2
(1 + eα) Λ(α) = 10 ln(E[eαBi^ ]) = 10{ln(1 + eα) − ln(2)}.
(c) The maximum is found by setting the derivative equal to zero:
0 =
d dα [Λ(α)^ −^ αc] = Λ
′(α) − c = 0 , where Λ′(α) = 10(1 + eα)− (^1) eα.
Consequently,
10(1 + eα)−^1 eα^ = c =⇒ 10 eα^ = c(1 + eα) =⇒ (10 − c)eα^ = c ,
from which we conclude that α∗^ = α∗(c) = ln(c) − ln(10 − c) for c < 10. Note that α∗^ = Λ(α∗) = 0 when if c = 5. The upper bound becomes trivial: P (X > 5) ≤ eΛ(α∗)−^5 α∗ = 1. However, this is the trivial case where c is equal to the mean, c = E[X] = 5. In general,
Λ(α∗) = 10[ln{1 + exp(ln(c) − ln(10 − c))} − ln(2)]
= 10
[
ln
c 10 − c
− ln(2)
]
[
ln
2 +^
c 2(10 − c)
)]
So,
P (X > c) ≤ exp
[
ln
2 +^
c 2(10 − c)
− ln
c 10 − c
· c
]
( 12 + 1210 c−c )^10 ( 10 c−c )c^
