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APPM 5570: Mathematical Statistics Solutions to Exam I Review Problems (21-27), Fall 2018
- The pdf for the P areto(γ) distribution is
f (x) =
γ (1 + x)γ+^
I(0,∞)(x).
The cdf is F (x) =
∫ (^) x
−∞
f (u) du =
∫ (^) x
0
γ (1 + u)γ+^
du = 1 −
(1 + x)γ for x > 0. The cdf for Yn is Fn(y) = P (Yn ≤ y) = P (n ln(X(1) + 1) ≤ y)
= P (X(1) ≤ ey/n^ − 1). Note that the cdf for the minimum is
FX(1) (x) = P (X(1) ≤ x) = 1 − P (X(1) > x)
= 1 − P (X 1 > x, X 2 > x,... , Xn > x)
iid = 1 − [P (X 1 > n)]n
[ 1 −
( 1 − (^) (1+^1 x)γ
)]n
= 1 − (^) (1+^1 x)nγ.
(So X(1) ∼ P areto(nγ).) Returning to the cdf for Yn:
Fn(y) = P (X(1) ≤ ey/n^ − 1)
= FX(1)(ey/n−1)
= 1 − (^) (1+ey/n^1 −1)nγ
= 1 − e−γy.
So,
nlim→∞ Fn(y) = lim n→∞ 1 −^ e−γy^ no n^ =^1 −^ e−γy. Thus, we have that Yn →d Y where Y ∼ exp(rate = γ).
- The pdf for the exponential distribution with rate λ is
f (x) = λe−λx^ I(0,∞)(x).
The cdf is F (x) =
∫ (^) x
−∞
f (u) du =
∫ (^) x
0
λe−λx^ dx = 1 − e−λx.
The cdf for the minimum is
FX(1) (x) = P (X(1) ≤ x) = 1 − P (X(1) > x)
= 1 − P (X 1 > x, X 2 > x,... , Xn > x)
iid = 1 − [P (X 1 > n)]n
= 1 − [1 − (1 − e−λx)]n
= 1 − e−nλx
which is the cdf for the exponential distribution with rate n. So, the cdf for Yn is Fn(y) = P (Yn ≤ y) = P (nX(1) ≤ y)
= P (X(1) ≤ y/n)
= 1 − e−nλ(y/n)
= 1 − e−λy. Once again, this is constant in n so
nlim→∞ Fn(y) = lim n→∞ 1 −^ e−λy^ no n^ =^1 −^ e−λy.
Thus, we have that Yn →d Y
where Y ∼ exp(rate = λ).
- (a) Let ε > 0. Note that
P (|(Xn + Yn) − (a + b)| > ε) = P (|(Xn + a) + (Yn + b)| > ε)
≤ P (|Xn − a| + |Yn − b| > ε)
cdf for Xn and let F be the cdf for X. All cdfs here are identical. They are not “moving in n”. So, we have nlim→∞ Fn(x) = lim n→∞ F^5 (x) =^ F^5 (x) =^ F^ (x)
On the other hand, note that
P (|Xn − X| > 2) = P (Xn = 1, X = −1) + P (Xn = − 1 , X = 1) indep =
This does not go to zero as n → ∞. Thus, we have exhibited an ε (ε = 2) so that
nlim→∞ P^ (|Xn^ −^ X|^ > ε)^6 →^0
and so we do not have convergence in probability of Xn to X.
- The exponential distribution with rate 1 has mean μ = 1 and variance σ^2 = 1. The sample mean for 100 values therefore has mean 1 and variance σ^2 /n = 1/100.
(a)
P (X > 1 .28) = P
( X − μ σ/
n
) ≈ P (Z > 2 .8)
where Z ∼ N (0, 1),by the CLT. From a z-table, we see that
P (X > 1 .28) ≈ 1 − 0 .9974 = 0. 0026.
(Note: I did not have a z-table with me when writing these solutions so don’t be con- cerned if your answer differs slightly.) (b) This problem should have asked for you to write down the asymptotic distribution of ∑ n i=1 Xi. As written, there is no asymptotic anything. In fact, we know that (^100) ∑
i=
Xi ∼ Γ(100, 1).
The CLT says that is should be approximately distributed as a N (100, 100) since that sum has mean 100 and variance 100. For this particular α and β combination, the gamma pdf looks like a nice symmetric bell curve. In fact, it is plotted in black in Figure 1 along with the N (100, 100) pdf in red. The CLT in action!!! If the problem had asked for the asymptotic distribution of
∑n i=1 Xi, the answer would be (^) n ∑
i=
Xi
asymp ∼ N (nμ, nσ^2 ) = N (n, n)
since μ and σ^2 are both 1.
Figure 1: Figure for 25b
- Let’s go for the joint cdf of the minimum and the maximum
FX(1),X(n) (x, y) = P (X(1) ≤ x, X(n) ≤ y).
It is not clear how to write this in terms of the individual Xi. Consider instead the relationship
P (X(n) ≤ y) = P (X(1) ≤ x, X(n) ≤ y) + P (X(1) > x, X(n) ≤ y). (1)
We know how to write out the term on the left-hand side. The first term on the right-hand side is what we want to compute. As for the final term,
P (X(1) > x, X(n) ≤ y),
note that this is zero if x ≥ y. (In this case, P (X(1) ≤ x, X(n) ≤ y) = P (X(n) ≤ y) and (1) gives us only P (X(n) ≤ y) = P (X(n) ≤ y) which is both true and uninteresting! So, we consider the case that x < y. Note then that
P (X(1) > x, X(n) ≤ y) = P (x < X 1 ≤ y, x < X 2 ≤ y,... , x < Xn ≤ y)
iid = [P (x < X 1 ≤^ y)]n
= [F (y) − F (x)]n.
Thus, from (1), we have that
FX(1),X(n) (x, y) = P (X(1) ≤ x, X(n) ≤ y)
= P (X(n) ≤ y) − P (X(1) > x, X(n) ≤ y)
= [F (y)]n^ − [F (y) − F (x)]n.
