By theorem B, chapter 6, "Mathematical Statistics and Data Analysis", 2nd edition, John Rice, page 181, which states that the distribution of $(n−12)S2- σ$ is a chi-square distribution with $n − 1$ degrees of freedom.

Hence

( (n − 1)S2 ) Var -----2---- = V ar(tn−1) σ

Since $nσ−21-$ is not random, then applying the property that $V ar(c X ) = c2Var (X )$ when $c$ is not random to the above, where in this case $c = (n−1) σ2$ and rearranging, we obtain

( 2) σ4 V ar S = -------2V ar (tn− 1) (n − 1)

However, $1 −n −n−1 1 I found V ar(tn− 1)(2fr)om Chi-2square′′moment ge′ner2ation function. Since M (tn) = (1− 2t)2 , then M (tn−1) = (1 − 2t) 2 and V ar(tn−1) = 2(n − 1) then Var(tn−1) = E t − E (t) = M (0)− [M (0)] which comes out to 2(n− 1)$ , hence

4 V ar(S2 ) = 2--σ--- n − 1

Let

∑n Sn = 1- |Xi | ni=1

Find moment generation function

To ﬁnd $( ) M |X| t n$ , and noting that $μ = 0$ and $σ = 1$ we obtain²

( ) ∫ t 1 ∞ t|x| −x2 M |X| -- = √---- e n e 2 dx n 2π −∞

Due to symmetry of normal distribution and since $|x|$ is positive always the above can be written as³

Hence

[ t2-( ( t ) )]n MSn (t) = e 2n2 1 + erf √---- 2n

The limit of the above as $n → ∞$ is $√ 2- e πt$ . Therefore

√ 2- MSn (t) = e πt

We see now that $′ ∘ -2 E (Sn ) = M (0) = π$ and $2 ′′ 2 E (S n) = M (0) = π$ , therefore $2 (∘ -2)2 V ar(Sn) = π − π = 0.$ (this means all sums add to same value for large $n$ , did I make a mistake? I did not expect this). Hence

( ∘ -- ) in distribution 2 Sn → N -,0 π

For pivotal term use $(n−1)s2 2 σ2 ˜ χ(n−1)$ , where $2 s$ is sample variance $2 σ$ is population variance, and hence we write (following class notes on 10/29/07) the conﬁdence interval as

P [− zp < 𝜃 < zp] = 1 − α

Where from table A7, $zp = 1.96$ for normal r.v. at $95%$ and Where $(n−1)2s22 𝜃 = (nσ−21)s2 --σ21-1$

Hence the C.I. becomes

Where the sample variance $2 -1--∑n ( ¯ )2 s2 = n−1 i=1 Xi − X$ , and $2 -1--∑n ( ¯ )2 s1 = n−1 i=1 Yi − Y$

For $95%$ conﬁdence, $α = 0.05$ . Hence the the ﬁnal answer for the C.I. is

|--[-------2-----2--------2]--------| | s1 σ1- s1 | |P − 1.96s22 < σ22 < 1.96s22 = 0.95 | -------------------------------------

Not sure what more I can do with the above so I think I will stop here.

First ﬁnd the joint density of $X, Y$ . Since $X, Y$ are independent, then the joint density $fX,Y (x, y)$ $= fX (x) fY (y )$ over $− ∞ < x < ∞$ and $y > 0$

But $√-1-- −(x−2σ2μ)2 fX (x) = 2πσ e$ and $−λy fY (y ) = λe$ , hence the joint density is (after substituting for $μ = 0,σ2 = 2,λ = 1$ is

|-------------------------------------------------------| | -1√--- −x42 −y | |fX,Y (x, y) = 2 πe e − ∞ < x < ∞, y > 0| ---------------------------------------------------------

Now Let $X-- Z = √Y-$ , and let $U = Y$

Hence

f (z,u) = |J|f (z,u) Z,U X,Y

(1)

Where

| | √ -- √ -- |J−1| = Y = U

Hence, from (1) and substitute $√ -- X = Z U$ and $Y = U$ , we obtain $\text{[math]}$

√ -- 1 −z2u −u fZ,U (z,u) = u2√-π-e 4 e

Hence the marginal density

|--------∫-∞---------------| fZ (z) = fZ,U (z, u) du | ----------0-----------------

Then

Now Gamma distribution is $λα α−1 − λw f (w ) = Γ (α-)w e$ , hence if we replace $z2 λ = 1 + 4-$ and $3 α = 2,$ then we have

To simplify further,

(3 ) f (z) = -1√----Γ--2---- Z 2 π( z2) 32 1 + 4

But $( ) √ π Γ 32 = -2-$ , hence

|----------------------3| | 1( z2) − 2| |fZ (z ) = -- 1 + --- | ----------4-------4------

Hence the pdf of $X√-- Y$ is

3 1( x2 )− 2 fX,Y (x,y ) = -- 1 + --- 4 4y

To verify this is a pdf, I integrate it from $− ∞$ to $+ ∞$ to see if I get 1:

Here is a plot of the distribution

Another attempt at problem (2)

By deﬁnition, the CDF of $| | | ¯Xn |$ is

Since $μ = 0,σ = 1$ we obtain

(| | ) ( ¯ ) F ¯ |X ¯n | = c = P --−√ c-< -Xn√---< --c√--- |Xn| 1∕ n 1∕ n 1∕ n

(2)

Now I need to combine (1) and (2). I am not sure how.

But central limit theorem tells us that as $n$ gets large, the distribution of the sample mean $X¯n$ approach normal distribution with mean $μ$ and variance $σ2 n$ , hence $( ) ¯ in distribution σ2 Xn → N μ, n$ , hence the above becomes

F |X¯n |(||X ¯n || = c) ≃ Φ (c√n-) − Φ (− c√n-)