Download Solutions to Problem Set #12 in ECE 313, University of Illinois, Spring 2003 and more Assignments Statistics in PDF only on Docsity!
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1.(a) In the figure below, I use • to denote a probability mass of 1/12 and • to denote a probability mass of 1/6.
2
1
3
1 2 3 4 5
4
(a) p X (0) = 1/6 + 1/12 = 1/4, p X (1) = 1/3, p X (3) = 1/4, p X (5) = 1/6. p Y (–1) = 2×1/12 + 1/6 = 1/3, p Y (3) = 1/3, p Y (4) = 1/3. (b) Since, for example, p X (3)p Y (3) = (1/3)×(1/4) =1/12 ≠ p X , Y (3,3) = 0, the random variables are not independent. We can also see this via the eyeball test: there are empty spots in the grid. (c) P{ X ≤ Y } = p X , Y (0,3) + p X , Y (1,3) + p X , Y (1,4) + p X , Y (3,4) = 1/2. P{ X + Y ≤ 8} = 1 – p X , Y (5,4) = 11/12. (d) Given Y = 3, X takes on values 0, 1 and 5 with conditional probabilities (1/6)/(1/3), (1/12)/(1/3), and (1/12)/(1/3), that is, 1/2, 1/4 and 1/4 respectively. Hence, E[ X | Y =3] = 0•(1/2) + 1•(1/4) + 5•(1/4) = 3/2, and E[ X^2 | Y =3] = 0^2 •(1/2) + 1^2 •(1/4) + 5^2 •(1/4) = 26/4, giving var( X | Y =3) = 26/4 – 9/4 = 17/4.
2.(a),(b) The joint pdf is nonzero on the quarter-plane shown. For v > 0, the pdf of Y is given by
f Y (v) = ∫
u=–v
v c•(v^2 –u^2 )•exp(–v)du = c•[v^2 u–u^3 /3]•exp(–v)
v
–v
= c•(4/3)•v^3 •exp(–v). We recognize this as a
gamma pdf with parameters (4,1) which gives us that c•(4/3) = 1/Γ(4) = 1/3! and thus c = 1/8.
For u > 0, f X (u) = ∫
v=u
∞ (1/8)•(v^2 –u^2 )•exp(–v)dv = (1/8)•(–v^2 –2v – 2 + u^2 ))•exp(–v)
u
= (1/4)•(1+u)•exp(–u)
while for u < 0, the integral is from v = -u to ∞. Thus, we get f X (u) = (1/4)•(1+|u|)•exp(–|u|). (c) The pdf of X is symmetric and the integral of u•(1/4)•(1+u)•exp(–u) from 0 to ∞ is finite (what’s that got to do with it??) Hence, E[ X ] = 0.
3.(a) The pdf is nonzero in the shaded region shown.
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u
v
(b) f X (u) = ∫
v=u
v=∞ 2 exp (–u–v) dv = 2•exp(–2u) if u > 0, and 0 otherwise.
f Y (v) = ∫
u=
u=v 2 exp (–u–v) du = 2•exp(–v) – 2•exp (–2v) if v > 0, and 0 otherwise.
(c) The random variables are not independent: f X (u)f Y (v) ≠ f X , Y (u,v).
(d) P{ Y > 3 X } = ∫
u=
v=3u
v=∞
2exp(–u–v)dvdu = ∫
u=
2exp(–4u)du = 1/2.
(e) P{ X + Y < α} = ∫
u=
α/
v=u
v=α–u
2e–u–vdvdu = ∫
u=
u=α/ 2e–u[e–u–e–α+u]du = 1 – (1 + α)exp(–α) for α > 0.
The probability is 0 for α ≤ 0.
(f) f X + Y (α) = ddα[1 – (1 + α)(exp(–α)] = α•exp (–α) for α > 0, and 0 otherwise. This is a gamma density
with parameters (2,1).
1−α
α
v
u
v
u
v
u
4. [“Look, Ma! No integrations!”] The joint pdf has values as shown in the two triangular regions. (a) f X (α) = area of cross–section at α = sum of areas of two rectangles = (1/2)(1 – α) + (3/2)α = (1/2) + α, for 0 ≤ α ≤ 1, and 0 if α < 0 or α > 1. (b) P{ X + Y ≤ 3/2} = 1 – P{ X + Y > 3/2} = 1 – volume above heavy line in middle figure above = 1 – (1/2)×(1/2)×(1/2)×(3/2) = 13/16. (c) P{ X^2 + Y^2 ≥ 1} = Volume outside heavy curve in right–hand figure above = (1 – π/4)×(3/2) = 3/2 – 3π/8. (Alternatively, volume inside the curve is (1/2)^2 + (π/4 – 1/2)×3/2 = 3π/8 – 1/2)
5.(a) Since X and Y are independent random variables, their joint pdf is the product of their (marginal) pdfs:
f X , Y (u,v) = (1/σ 2 π)•exp(–u^2 /2σ^2 )•(1/σ 2 π)•exp(–v^2 /2σ^2 ) = (1/2πσ^2 )•exp(–(u^2 +v^2 )/2σ^2 ) for all u, v. (b) The region is the exterior of the circle of radius α as indicated in the left hand sketch below.
