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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 if1

if1 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,

if 0 ≤ u ≤1 if1 2< u ≤ 1

(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�

Xi

Since (Xm+1, . . . , Xn) is independent of (X1, . . . , Xm), it is also independent of Sm. Thus ∀s ∈ R: P(Sn ≤ t|Sm = s) = P = P �
i=m+1

= P

n

Xi ≤ t i=m+1� � Xi ≤ t − s����� Sm = sn n

i=m+1�
n

= m + 0 −m n· n = 0

Xi

2e−y/2� 2y 2e−x/2dxdy
1 1

1

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 .

Therefore EM2= E[E[M2|N]] = E[25N2+25 3N] = 25 · (102 + 10) + 25 3· 10 = 8500 3.

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.

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

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.

Remark: We can also compute directly the joint density of (R, Θ) and conclude that (1) R is inde-pendent of Θ, and (2) Θ has a uniform distribution on [0, 2π).

fR(r) = re− 1 2r2

For r ≥ 0, θ ∈ [0, π/2), because tan(·) is monotone on [0, π/2),

P(R ≤ r, 0 Θ ≤ θ) = P(X2+ Y2≤ r2, tan Θ tan θ) = P(X2+ Y2≤ r2, Y/X ≤ tan θ)

∂x ∂x

det(J) = r1

J = ∂r1∂y

∂θ1
∂y

= cos θ1
∂r1 ∂θ1 1 1

fR,Θ(r, θ) = fR(r)fΘ(θ)

where fR(r) = re− 1 2r21{r>0} and fΘ(θ) = 1{0θ<2π}

fR(r)dr =

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
=

Problem 7. Addition of Gammas. The Gamma density with shape parameter α > 0 and scale parameter λ > 0 is

fα,λ(x) =λαxα1e−λx Γ(α) for x ≥ 0,
= 0 for x < 0.

fX+Y (z) =

−∞fX(u)fY (z − u)du

Γ(β)

z

=
=
uα−1(z − u)β−1 Γ(α)Γ(β)
=

λα+βe−λzzα+β−1

�0
1

tα−1(1 − t)β−1 Γ(α)Γ(β) dt
=
= Γ(α + β)
FX2(t) =

P(X2≤ t) = P(−√t √t ≤ X ≤√t)

√t 2πe−x2/2 dx = Φ( 1 √t) Φ(−√t).

=

(C) From part (a) and (b) and the fact that X2 iare i.i.d we get that

i=1 n X2 i∼ Gamma( i=1 n 2, 1 2) i.e.

Gamma(n 2, 1 2).

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