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Thus the joint density function of U and V is, by Theorem 2-4, c (v2> 3u 0 e u 0
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1 u 3
2>3 v 2>3
g(u, v)
(3 u
2>3 v 2>3)
64u, v
125v
otherwise v2 64u, otherwise v u2 125v
g(u, v)
Convolutions 2.23. Let X and Y be random variables having joint density function f (x, y). Prove that the density function of U X Y is
g(u) Method 1
Let U u x
f (v, u
v)dv
X Y, V X, where we have arbitrarily added the second equation. Corresponding to these we have y, v x or x v, y u v. The Jacobian of the transformation is given by 'x 'u 4 'y 'u 'x 'v 4 'y 'v 0 1 1 2 1
Thus by Theorem 2-4, page 42, the joint density function of U and V is g(u, v) f (v, u v)
It follows from (26), page 41, that the marginal density function of U is
g(u)
f (v, u
v) dv
CHAPTER 2 Random Variables and Probability Distributions
Method 2 The distribution function of U by x y u, i.e.,
Y is equal to the double integral of f (x, y) taken over the region defined
G(u)
f (x, y) dx dy
Since the region is below the line x
u, as indicated by the shading in Fig. 2-16, we see that
G(u)
f (x, y) dyR dx
Fig. 2-16
The density function of U is the derivative of G (u) with respect to u and is given by
g(u)
3 `f (x, u
x) dx
using (12) first on the x integral and then on the y integral.
2.24. Work Problem 2.23 if X and Y are independent random variables having density functions f1(x), f2( y), respectively.
In this case the joint density function is f (x, y) of U X Y is
f 1(x) f2( y), so that by Problem 2.23 the density function 3 f1(v) f2(u v)dv f1 * f2
g(u) which is the convolution of f1 and f2.
2.25. If X and Y are independent random variables having density functions f1(x) e 2e 0
x x X
0 0 Y.
f2 (y)
3e 0
find the density function of their sum, U
By Problem 2.24 the required density function is the convolution of f1 and f2 and is given by g(u) In the integrand f1 vanishes when v
f1 * f2
f1(v) f2(u
v) dv
0 and f2 vanishes when v
2v)(3e 3(u v)) dv u
u. Hence
g(u)
30 (2e 6e
v 30 e dv
3u (e u
e3u)
CHAPTER 2 Random Variables and Probability Distributions
if u
0 and g(u)
0 if u
Check:
g(u) du
6 3 (e
3u) du
2.26. Prove that f1 * f2
We have
f2 * f1 (Property 1, page 43).
1 3
f1 * f2 Letting w u v so that v f1 * f2 u
f1(v) f2(u
v) dv
w, dv
dw, we obtain
f1(u
w) f2(w)( dw)
f2(w)f1 (u
w) dw
f2 * f1
Conditional distributions 2.27. Find (a) f ( y u 2), (b) P(Y
2) for the distribution of Problem 2.8.
(a) Using the results in Problems 2.8 and 2.9, we have f ( y u x) so that with x 2 f ( y u 2) (b) (4 y)>42 11>21 4 22 y f (x, y) f1(x) (2x y)>42 f1(x)
1u X
f (1 u 2)
5 22
2.28. If X and Y have the joint density function
f (x, y) find (a) f ( y u x), (b) P(Y (a) For 0 x 1, f1(x)
e4 0
3 1 2
0 x 1, 0 otherwise
dx).
1 3 30 4
xy dy 3 3 0
3 4 0
x 2 y 1
f ( y u x)
f (x, y) f1(x)
4xy 2x
other y
For other values of x, f ( y u x) is not defined. (b) P(Y
31>2
f ( y u 1) dy 2
31> 2
9 16
2.29. The joint density function of the random variables X and Y is given by f (x, y) e 8xy 0 0 x 1, 0 otherwise y x
Find (a) the marginal density of X, (b) the marginal density of Y, (c) the conditional density of X, (d) the conditional density of Y.
The region over which f (x, y) is different from zero is shown shaded in Fig. 2-17.
CHAPTER 2 Random Variables and Probability Distributions
Fig. 2-17
(a) To obtain the marginal density of X, we fix x and integrate with respect to y from 0 to x as indicated by the vertical strip in Fig. 2-17. The result is
f1(x) for 0 x 1. For all other values of x, f1 (x)
3y 08xy dy 0.
4x 3
(b) Similarly, the marginal density of Y is obtained by fixing y and integrating with respect to x from x as indicated by the horizontal strip in Fig. 2-17. The result is, for 0 y 1,
y to x
f2 ( y) For all other values of y, f2 ( y) 0.
8xy dx
4y(1
y 2)
(c) The conditional density function of X is, for 0 f1(x u y) f (x, y) f2 (y)
1, 2x>(1 0 y 2) 0. y x other x 1
The conditional density function is not defined when f2( y) (d) The conditional density function of Y is, for 0 f2( y u x) f (x, y) f1(x) x e 1,
2y>x 2 0
0 y other y 0.
The conditional density function is not defined when f1(x)
1 1 3 30 4x dx 1
Check:
30 f1(x) dx
30 f2( y) dy 2x y2 dx 1
30 4y(1 1
y 2) dy
3y f1(x u y) dx 30 f2( y u x) dy
3y 1
x 2y 30 x 2 dy
2.30. Determine whether the random variables of Problem 2.29 are independent.
In the shaded region of Fig. 2-17, f (x, y) 8xy, f1(x) 4x3, f2( y) 4y (1 y2). Hence f (x, y) 2 f1(x) f2( y), and thus X and Y are dependent. It should be noted that it does not follow from f (x, y) 8xy that f (x, y) can be expressed as a function of x alone times a function of y alone. This is because the restriction 0 y x occurs. If this were replaced by some restriction on y not depending on x (as in Problem 2.21), such a conclusion would be valid.
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