The Uniform Distribution and Its Relation to the Standard Normal Distribution
As we have seen before in Chapter 1, the uniform [0,1] distribution has density
f (x) = 1 for 0 < x < 1, f (x) = 0 elsewhere.
More generally, the uniform [a, b] distribution (denoted by U[a, b]) has density
where U and U2 are independent U[0, 1] distributed. Then Xj and X2 are
independent, standard normally distributed. This method is called the Box – Muller algorithm.
The x2 distribution is a special case of a Gamma distribution. The density of the Gamma distribution is
This distribution is denoted by Г (а, в). Thus, the x2 distribution is a Gamma distribution with a = n/2 and в = 2.
The Gamma distribution has momentgenerating function
тг(а, в)(0 = [1 – ві]a, t < 1/в (4.44)
and characteristic function (рГ(а, в'() = [1 – в • i • t]—a – Therefore, the Г(а, в) distribution has expectation ав and variance ав2.
The Г(а, в) distribution with a = 1 is called the exponential distribution.
1. Derive (4.2).
2. Derive (4.4) and (4.5) directly from (4.3).
3. Derive (4.4) and (4.5) from the momentgenerating function (4.6).
4. Derive (4.8), (4.9), and (4.10).
5. If X is discrete and Y = g(x), do we need to require that g be Borel measurable?
6. Prove the last equality in (4.14).
7. Prove Theorem 4.1, using characteristic functions.
8. Prove that (4.25) holds for all four cases in (4.24).
9. Let X be a random variable with continuous distribution function F(x). Derive the distribution of Y = F(X).
10. The standard normal distribution has density f(x) = exp(x2/2)Д/2л, x є К. Let X1 and X2 be independent random drawings from the standard normal distribution involved, and let Y1 = X1 + X2, Y2 = X1 — X2. Derive the joint density h(y1, y2) of Y1 and Y2, and show that Y1 and Y2 are independent. Hint: Use Theorem 4.3.
11. The exponential distribution has density f (x) = 9—1 exp(x /9) if x > 0 and f (x) = 0 if x < 0, where 9 > 0 is a constant. Let X1 and X2 be independent random drawings from the exponential distribution involved and let Y1 = X1 + X2, Y2 = X1 — X2. Derive the joint density h( y1, y2) of Y1 and Y2. Hints: Determine first the support {(y1, y2)T є К2 : h(y1, y2) > 0} of h(y1, y2) and then use Theorem 4.3.
12. Let X ~ N(0, 1). Derive E[X2k] for k = 2, 3, 4, using the momentgenerating function.
13. Let X1, X2,…, Xn be independent, standard normally distributed. Show that (1/»E"j=1 Xj is standard normally distributed.
14. Prove (4.31).
15. Show that for t < 1 /2, (4.33) is the momentgenerating function of (4.34).
16. Explain why the momentgenerating function of the tn distribution does not exist.
17. Prove (4.36).
18. Prove (4.37).
19. Let X1, X2,…, Xn be independent, standard Cauchy distributed. Show that (1/n)Ej=1 Xj is standard Cauchy distributed.
20. The class of standard stable distributions consists of distributions with char
acteristic functions of the type p(t) = exp(—t a/a), where а є (0, 2]. Note that the standard normal distribution is stable with а = 2, and the standard Cauchy distribution is stable with а = 1. Show that for a random sample X1, X2, Xn from a standard stable distribution with parameter a, the ran
dom variable Yn = n—1/a ‘Yl’j=1 Xj has the same standard stable distribution (this is the reason for calling these distributions stable).
21. Let X and Y be independent, standard normally distributed. Derive the distribution of X/Y.
22. Derive the characteristic function of the distribution with density exp(—x )/2, —ж < x < ж.
23. Explain why the momentgenerating function of the Fm, n distribution does not exist.
24. Prove (4.44).
25. Show that if U and U2 are independent U[0, 1] distributed, then X1 and X2 in (4.43) are independent, standard normally distributed.
26. If X and Y are independent Г(1, 1) distributed, what is the distribution of XY ?

4. A. Tedious Derivations Derivation of (4.38):
пГ((п + 1)/2) f x2/n,
^ППГ(п/2) J (1 + x2/n)(n+1)/2 dX
пГ((п + 1)/2) Г 1 + x2/n d
4ПЛГ{п/2) ] (1 + x2/п)(п+1)/2 dX
пГ((п + 1)/2W 1 d
^пЛГ(п/2) J (1 + x2/п)(п+1)/2
пГ((п – 1)/2 + 1) Г(п/2 – 1) п
——————————————— п = ——— .
Г(п/2) Г((п – 1)/2) п – 2
In this derivation I have used (4.36) and the fact that
1 = j h^2(x)dx
_ Г((п – 1)/2) f _________ 1________ dx
Мп – 2)пГ((п – 2)/2) J (1 + x2 /(п – 2))(п1)/2 x
= Г((п – 1)/2W 1 d
^ЛГ((п – 2)/2) J (1 + x2)(пі)/2
Derivation of (4.40): For m > 0, we have
m
2П J exp(
m


1 f 1 f
— I exp[(1 + i ■ x)t]dt +—— I exp[(1 — i ■ x)t]dt
2n J 2n J
00
2n (1 — i ■ x)
1 exp(—m)
n (1 + x2) n (1 + x2)
Letting m ^ to, we find that (4.40) follows. Derivation of (4.41):
hm, n (x) Hm, n (x)
yn/2 1 exp(— y/2) .
X T(n/2)2n/1 d
mm/2 xm/2—1
nm/2 Y(m/2)Y(n/2) 2m/2+n/2
TO
Xf y",2+"2—[1+mx/n] y/2) dy
0
mm /2 xm/2 1
nm/2 Y(m/2)Y(n/2) [1 + m ■ x/n]m/2+n/2
TO
x j zm/2+n/2—X exp (—z) dz
mm/2 Y(m/2 + n/2) xm/2—1
nm/2 Y(m/2)Y(n/2) [1 + m ■ x/n]m/2+n/2 ’
Derivation of (4.42): It follows from (4.41) that
hence, if k < n /2, then
TO
J xk hm, n (x)dx 0
TO
mm/2 Tim/2 + n/2) f xm/2 + k1
—_______________ і _________________ dx
nm/1 T(m/2)T(n/2) J (1 + m ■ x/n)m/2 + n/2
0
TO
)k Tm/2 + nmf_______ xm + *<r–1_____
T(m/2)T(n/2)j (1 + x)(m + Ml2 + (n – ЖГ
0
T(m/2 + k)T(n/2 – к)
T(m /2)T(n/2)
nk=0(m /2+j )
Ytj=1(n/2 – j )’
where the last equality follows from the fact that, by (4.36), T(a + k) = T(a) nk=0(« + J) for a > 0. Thus,
= to if n < 4.
The results in (4.42) follow now from (4.46) and (4.47).
For notational convenience I will prove Theorem 4.4 for the case k = 2 only. First note that the distribution of Y is absolutely continuous because, for arbitrary Borel sets B in K2,
P[Y є B] = P [G(X) є B] = P [X є G1(B)] = j f(x)dx.
G1(B)
If B has Lebesgue measure zero, then, because G is a onetoone mapping, the Borel set A = G1(B) has Lebesgue measure zero. Therefore, Y has density
h(y), for instance, and thus for arbitrary Borel sets B in K2,
Choose a fixed y0 = (y0>1, y0,2)T in the support G(K2) of Y such that x0 = G1(y0) is a continuity point of the density f of X and y0 is a continuity point of the density h of Y. Let Y(5b «2) = [y0,1, У0,1 + «1] x [70,2, У0,2 + «2] for
some positive numbers 51 and 52. Then, with X the Lebesgue measure
P [Y є Y(51,52)]
and similarly,
P [Y є Y(5b 52)] >( inf f(G1(y))) X(G1(Y(51, 52))).
yeY(31,«2) /
(4.49)
It follows now from (4.48) and (4.49) that P [Y є Y(51,52)]
51 52
It remains to show that the latter limit is equal to det[ J(y0)] .
If we let G1 (y) = (g* ( y), gf( y))T, it follows from the mean value theorem that for each element g*(y) there exists a Xj є [0, 1] depending on y and y0 such that g*(y) = g*У0) + JjУ0 + Xj(y – y0))(y – У0), where Jj(y) is the jth row of J(y). Thus, writing
D ( )_ ( J1(y0 + X1(y – y0)) – J1(y0)£
0(Л~ J2(y0 + X2(y – y0)) – J2(y0) J
= J0(y) – J(y0), (4.51)
for instance, we have G1(y) = G1(y0) + J(y0)(y – y0) + D0(y)(y – y0). Now, put A = J(y0)1 and b = y0 – J(y0)1 G1(y0). Then,
G1(y) = A1(y – b) + D0(y)(y – y0);
hence,
G1(Y(«i,«2)) = {x є К2:x
= A1(y – b) + D0(y)(y – yo), y є Y(5i, «2)}
(4.53)
The matrix A maps the set (4.53) onto A[G1(Y(5i,52))]
= {x є К2 : x = y – b + A ■ D0(y)(y – У0), y є Y(«i, «2)},
(4.54)
def
where for arbitrary Borel sets B conformable with a matrix A, A [ B ] = {x : x = Ay, y є B}. Because the Lebesgue measure is invariant for location shifts (i. e., the vector b in (4.54)), it follows that
X (A[G1(Y(«i, «2))])
= X ({x є К2 : x = y + A ■ D0(y)(y – y0), y є Y(«i, «2)}) –
(4.55)
Observe from (4.51) that
A ■ Do(y) = J(yo)1 Do(y) = J(yo)1 Jo(y) – I2 (4.56)
and
lim J(yo)1 J o(y) = I2 (4.57)
У^Уо
Then
X (A[G1(Y(«i,«2))])
= X ({x є К2 : x = yo + J(yo)1 Jo(у)(У – Уо), y є Y(«i, «2)}) –
(4.58)
It can be shown, using (4.57), that
X (A[G1(Y(«i,«2))]) lim lim —1 — 1
5i40 «240 X (Y(«i,«2))
Recall from Appendix I that the matrix A can be written as A = QDU, where Q is an orthogonal matrix, D is a diagonal matrix, and U is an uppertriangular matrix with diagonal elements all equal to 1. Let B = (0, 1) x (0, 1). Then it is not hard to verify in the 2 x 2 case that U maps B onto a parallelogram U[B] with the same area as B; hence, X(U[B]) = X(B) = 1. Consequently, the Lebesgue measure ofthe rectangle D[B] is the same as the Lebesgue measure of the set D[U [ B]]. Moreover, an orthogonal matrix rotates a set of points around the origin, leaving all the angles and distances the same. Therefore, the set A [B]
has the same Lebesgue measure as the rectangle D[B] :k(A[B]) = k(D[B ]) = det[D] = det[A]. Along the same lines, the following more general result can be shown:
Lemma 4.B.1: For a k x k matrix A and a Borel set B in Kk, к(A[B]) = det[A] к(B), where к is the Lebesgue measure on the Borel sets in Kk.
Thus, (4.59) now becomes
limlim к (A[G"(V№.fe))])
hence,
1
det[ A]
det[ A1] = det[J(^0)] (4.60)
Theorem 4.4 follows now from (4.50) and (4.60).
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