SOLUTION FOR HOMEWORK 10, STAT 4351 Welcome to your 10th ...

Here o λ (1) → 0 as λ → ∞ is a standard mathematical notation “o-small of 1”, and we get
M Z (t) → e −tλ1/2 +tλ 1/2 +(1/2)t 2 = e t2 /2 , λ → ∞.
Because e t2 /2 is the mgf of a standard normal RV, we proved a remarkable result (a version
of the central limit theorem) that Poisson RV converges to Normal RV in distribution as
λ → ∞.
8. Problem 6.45. First of all, let us recall that the exponent in a bivariate normal is
1 x − µ 1 ) 2
−
2(1 − ρ 2[
σ 2 1
− 2ρ x − µ 1 y − µ 2
+ (y − µ 2) 2 ]
.
σ 1 σ 2 σ2
2
Now, for the distribution at hand, the exponent is
−(1/102)[(x + 2) 2 − 2.8(x + 2)(y − 1) + 4(y − 1) 2 ].
(a). We need to equate the terms. Plainly we get µ 1 = −2, µ 2 = 1, plus three equations
which we should solve as a system:
1
2(1 − ρ 2 )σ 2 1
= 1/102,
2ρ
2(1 − ρ 2 )σ 1 σ 2
= 2.8
102 ,
1
2(1 − ρ 2 )σ 2 2
= 4/102.
Dividing first on the third we get σ 1 = 2σ 2 . The second one gives us 2(1−ρ 2 )σ2 2 /ρ = 102/2.8
and combining it with the third we get ρ = .7. Finally, we get σ 2 = 5 and σ 1 = 10.
(b) We know that (see Theorem 6.9 on page 221)
E(Y |X = x) = µ 2 + ρσ 2 σ −1
1 (x − µ 1).
Plug-in numbers and get the answer.
Similarly,
V ar(Y |X = x) = σ 2 2 (1 − ρ2 ).
Plug-in numbers and get the answer.
9. Problem 6.55. Here X ∼ Exp(θ = 120).
(a)
∫ 24
P(X ≤ 24) = (120) −1 e −t/120 dt = −e −t/120∣ ∣24
0
0 = 1 − e−24/120 = 1 − e −.2 .
(b)
∫ ∞
P(X > 180) = (120) −1 e −t/120 dt = −e −t/120∣ ∣∞
180
180 = e−3/2 .
3