The gamma density with shape parameter and scale parameter isf for
Solution.
(a) We first compute the conditional c.d.f., namely, for u1, u2 ∈ [0, 1], the probability P(U1 ≤u1, U2 ≤ u2|M ≤ 1/2).
P(U1 ≤ 1/2)P(U2 ≤ 1/2)
= min(u1, 1/2) · min(u2, 1/2)
1,
if 0 ≤ u1 ≤1
if 0 ≤ u1 ≤1 if1if1 2< u1 ≤ 1, 0 ≤ u2 ≤ 1 2< u1 ≤ 1,
1
2, 0 ≤ u2 ≤ 1 2, 1 2< u2 ≤ 1
The denominator is given by
P(M ≥ 1/2) = 1 − P(M < 1/2) = 1 − P(U1 < 1/2, U2 < 1/2) = 3/4
=�u −1 2u,
4, if 0 ≤ u ≤1 if1 2< u ≤ 1
1
h(u|M ≥ 1/2) = | 3, |
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(a) Let U = X and V = XY , then (X, Y ) → (U, V ) is a one-to-one transformation from A := {(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1} to B := {(u, v) : 0 ≤ u ≤ 1, 0 ≤ v ≤ u}. The Jacobian matrix is
J =� ∂u ∂y
2
(a) Given Sm = s, | Sn = | � | Xi = Sm + | i=m+1� | Xi = s + | i=m+1� |
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= m + 0 −m n· n = 0 |
Xi | � | |||||||
2e−y/2� 2y 2e−x/2dxdy
1 11
Solution. Let N be the total number of runners who arrive at a time between 0 and 10, and let T1, . . . , TN be their arrival times. By assumption, we have M =�N i=1(10 − Ti). Since the runners
• N follows a Poisson distribution with mean 10.
(b) Conditioning on N = n,
N
= nE[U2 1] + n(n − 1)E[U1U2] = 100 3n + 25n(n − 1) = 25n2 + 25 3n .
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Problem 6. Independent Gaussians. Suppose that X and Y are independent random variables each with the standard normal distribution. Let X = R cos Θ and Y = R sin Θ be the polar coordi-nate representation of the point (X, Y ), with the angular coordinate Θ chosen so that 0 ≤ Θ < 2π. (A) Find the density of Y/X. (Hint: connect this to the angle of the point (X, Y )).
are independent standard normal random variables. NOTE: You can either look up the relevant trig double-angle formulas or you can forget all the trig you ever knew and just learn Euler’s
eiθ= cos θ + i sin θ. |
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that Θ is uniformly distributed on [0, 2π). So R is also independent of 2Θ (mod 2π). Next, we show that 2Θ (mod 2π) and Θ have the same distribution, both of which are uniform over [0, 2π).
Firstly, they clearly have the same range,
2Θ(mod 2π) ∈ [0, 2π) | and |
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Since the joint distribution of a pair of independent random variables is exclusively determined by their marginals, it follows that
(R, 2Θ mod 2π)D= (R, Θ)
Combining all these facts, we finally get� � R cos(2Θ mod 2π), R sin(2Θ mod 2π))D= (R cos 2Θ, R sin 2Θ
( ˜X,˜Y ) = (R cos 2Θ, R sin 2Θ
Thus˜X and˜Y are independent, standard normal random variables, since X and Y are.
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fR(r) = re− |
For r ≥ 0, θ ∈ [0, π/2), because tan(·) is monotone on [0, π/2),
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∂x | ∂x |
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J = | ∂r1∂y |
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= | cos θ1 | � | |||||||
∂r1 | ∂θ1 | 1 1 | � |
fR,Θ(r, θ) = fR(r)fΘ(θ)
where fR(r) = re−
12r21{r>0} and fΘ(θ) = 1{0≤θ<2π}
fR(r)dr = | � ∞ |
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We thus conclude that R and Θ are independent, and Θ has a uniform distribution on [0, 2π). 7
(C) This immediately follows from (B) once we notice that,
= | R2cos2Θ − R2sin2Θ R | |
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= |
Problem 7. Addition of Gammas. The Gamma density with shape parameter α > 0 and scale parameter λ > 0 is
fα,λ(x) =λαxα−1e−λx Γ(α) for x ≥ 0,
= 0 for x < 0.
fX+Y (z) | = |
Γ(β)
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= | |||||||||
= | uα−1(z − u)β−1 Γ(α)Γ(β) | ||||||||
= |
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tα−1(1 − t)β−1 Γ(α)Γ(β) | dt | |||||
= | |||||||||
= | Γ(α + β) |
FX2(t) | = |
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= | |||||
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i=1 n� | X2 i∼ Gamma( i=1 n� | 2, |
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