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Material Type: Notes; Class: PROBAB MTH,ELEC ENG; Subject: STATISTICS; University: Iowa State University; Term: Unknown 1989;
Typology: Study notes
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Reading: G. R. Cooper & C. D. McGillem 2.6 - 2.
EE/STAT 322, #7 1
The PDF is given by
fX (x) =
τ e−^
(^1) τ x (^) x ≥ 0 0 otherwise
E(X) = τ , and σ x^2 = τ 2.
E(X) = ∫^0 ∞ f (x)xdx = ∫^0 ∞^1 τ e−^1 τ^ xxdx. By integrating by parts (∫^ xdy = xy| − ∫^ ydx), we get
E(X) = −
0 xde
− (^) τ^1 x
= −xe−^1 τ^ x|∞ 0 +
0 e
− (^) τ^1 xdx = τ
EE/STAT 322, #7 3
Example: The waiting time of a customer has an exponential distribution with a mean of 5 minutes. Then what is the probability that he will wait more than 10 minutes?
Solution: τ = 5, Pr(X > 10) = e−^10 /τ^ = e−^2 = 0. 1353.
If he already spent five minutes there, the chance that he needs to wait another 10 minutes is given by
Pr(X > 15 |X > 5) = e−^15 /τ^ /e−^5 /τ^ = Pr(X > 10) = 0. 1353.
EE/STAT 322, #7 7
Example: (Ex 2-7.3)
Three coins are tossed. Let H = {number of heads}.
Find E(H), σ H^2.
Solution: We have four possible outcomes:
H 0 = 0, with probability p 0 = 213 (^30 )^ = 1/ 8. H 1 = 1, with probability p 1 = 213 (^31 )^ = 3/ 8. H 2 = 2, with probability p 2 = 213 (^32 )^ = 3/ 8. H 3 = 3, with probability p 3 = 213 (^33 )^ = 1/ 8.
The PDF is p(H) = ∑^3 i=0 piδ(H − Hi).
E(H) = p 0 · 0 + p 1 · 1 + p 2 · 2 + p 3 · 3 = 12/8 = 1. 5.
E(H^2 ) = p 0 · 0 + p 1 · 12 + p 2 · 22 + p 3 · 32 = 24/8 = 3.
σ^2 H = E(H^2 ) − E(H)^2 = 3 − 1. 52 = 0. 75.
EE/STAT 322, #7 9
Example: (Problem 2-4.5, textbook) RV X has a pdf of the form
fX (x) = ax^2 0 < x ≤ 2 = ax 2 < x ≤ 3
Find (a) the value of a, (b) E(X), (c) Pr(2 < X ≤ 3). Solution:
(a) We need ∫^03 fX (x)dx = 1, so that ∫^02 ax^2 dx + ∫^23 ax = ax^3 / 3 |^20 + ax^2 / 2 |^32 = a[8/3 + 9/ 2 − 4 /2] = 1, → a = 6/ 31. (b) E(X) = ∫^02 xax^2 dx + ∫^23 xaxdx = 316 [4 + 9 − 8 /3] = 2. (c) Pr(2 < X ≤ 3) = ∫^23 fX (x)dx = ∫^23316 xdx = 15/ 31.
Example: (2-6.3) A current I with a Rayleigh PDF passes through a resistor with (R = 2π)Ω. E(I) = 2 A. Power dissipation: W = RI 2. (a) Find the mean of power dissipated E(W ). (b) Find Pr(W ≤ 12). (c) Pr(W > 72). Solution:
(a) E(I) = 2 = √π/ 2 σ → σ = 2√ 2 /π. E(W ) = RE(I^2 ) = R(2σ^2 ) = 2π 2 · 4 · 2 /π = 32. (b) Pr(W ≤ 12) = Pr(RI^2 ≤ 12) = Pr(I^2 < 1 .91) = Pr(I < 1 .382) = FI (1.382) = 1 − exp(−^1. 2382 σ 2 2 ) = 0. 3127. (c) Pr(W > 72) = Pr(I > 3 .385) = 1 − FI (3.385) = 0. 1054.
EE/STAT 322, #7 13
Example: (2-7.1) RV θ is uniformly distributed in (0, 2 π). Another RV X is given by X = cos(θ). (a) Find PDF of X; (b) Find E(X); (c) Find σ^2 x; (d) Find Pr(X > 0 .5). Solution:
(a) dx/dθ = − sin θ = −√ 1 − cos^2 θ = −√ 1 − x^2.
cos(θ )
x
x = cos θ has two roots (solutions) in (0, 2 π), given by θ 1 = cos−^1 x ∈ (0, π), and θ 2 = 2π − cos−^1 x ∈ (π, 2 π). EE/STAT 322, #7 15
At both θ 1 and θ 2 , |dx/dθ| = √ 1 − x^2.
fX (x) = √^21 f −^ (θ x) 2 = (^) π√ 11 − x 2 − 1 ≤ x ≤ 1 = 0 elsewhere
(b) E(X) = E[cos(θ)] = ∫^02 πcos(θ) (^21) π dθ = 0. (c) E(X^2 ) = X^2 = E[cos^2 θ] = ∫^02 πcos^2 (θ) (^21) π dθ = 12. σ^2 x = X^2 − X^2 = X^2 = 12.
