Chapter 6 Random Variables Section 6.2 Transforming and Combining Random Variables.
Function of Random Variables - people.ysu.edu
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Function of Random Variables
FRV - 1
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Functions of Random Variables
7.1 Introduction
Methods for finding distribution of function of one
or more random variables:
1. Distribution Function Technique
2. Transformation Technique
3. Moment Generating Function Technique
How to find the distribution of a random variable Y
that is a function of several random variables X1, X2,
โฆ, Xn that has a joint probability distribution?
๐ฆ = ๐ข(๐ฅ1, ๐ฅ2, โฆ, ๐ฅ๐)
7.2 Distribution Function Technique
For finding the probability density function with a
given joint probability density, the probability
density function of ๐ = ๐ข(๐1, ๐2, โฆ, ๐๐) can be
obtained by first finding the cumulative probability
or distribution function
F(y)= ๐ ๐ โค y = ๐(๐ข(๐1, ๐2, โฆ, ๐๐)
and then differentiate it to get the p.d.f.
๐ ๐ฆ =๐๐น(๐ฆ)
๐๐ฆ
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Example: Let X ~ U(0,1), and Y = ๐๐ , find ๐. ๐. ๐. of ๐.
G(y) = P(Y โค y)
= P(๐๐ โค y)
= P(X โค y1/n )
= F(y1/n )
= y1/n
g ๐ฆ = 1
๐๐ฆ
1๐โ1, 0 โค ๐ฆ โค 1
0, elswhere.
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Example: Let X has the following p.d.f.,
G(y) = P(Y โค y) = P(๐3 โค y)
= P(X โค y1/3 )
= 6๐ฅ 1 โ ๐ฅ ๐๐ฅy1/3
0
= 3y2/3 โ 2y, for 0< y <1
๐ ๐ฅ = 6๐ฅ 1 โ ๐ฅ , 0 < ๐ฅ < 1
0, elswhere
g ๐ฆ = 2(y โ1/3 โ 1), 0 < ๐ฆ < 10, elswhere.
find the p.d.f. of Y = ๐3.
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Example: Let X have a p.d.f. f(x) and Y = ๐2 , find ๐. ๐. ๐. of ๐.
G(y) = P(Y โค y)
= P(๐2 โค y)
= P(-y1/2 โค X โค y1/2)
= F(y1/2) - F(- y1/2)
g(y) = 1
2 ๐ฆ๐(y1/2) + f(โ y1/2)
Function of Random Variables
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Example: Let X have a p.d.f. f(x) and Y = |X|, find ๐. ๐. ๐. of ๐ when ๐ ~ ๐ 0,1 .
G(y) = P(Y โค y)
= P(|X|โค y)
= P(-y โค X โค y)
= F(y) - F(- y)
g(y) = ๐(y) + f(โy), for y > 0
when ๐ ~ ๐ 0,1 , g(y) = 2๐(y) , for y > 0
= 21
2๐๐โ
๐ฆ2
2
-y 0 y 8
Example: Let X have a p.d.f. ๐ ๐ฅ =1
๐๐โ
๐ฅ
๐, for ๐ฅ > 0,
find p. d. f. of ๐ = ln ๐ .
G(y) = P(Y โค y)
= P(lnX โค y)
= P(X โค e y )
= 1
๐๐โ
๐ฅ
๐ ๐๐ฅ๐๐ฆ
0
= - ๐โ๐ฅ
๐ ๐๐ฆ
0= 1 โ ๐โ
1
๐โ๐๐ฆ
g(y) = Gโฒ(y) = 1
๐๐๐ฆ๐โ
1
๐โ๐๐ฆ
, - < y <
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Example: If the joint density of X1 and X2 is given by
Find the probability of Y = X1 + X2
F(y) = P(Y โค y) = P(X1 + X2 โค y)
๐ ๐ฅ1, ๐ฅ2 = 6๐โ3๐ฅ1โ2๐ฅ2 , for ๐ฅ1 > 0, ๐ฅ2 > 0
0, elsewhere
= 1 + 2e-3y โ 3e-2y
F(y) = f (y) = 6(e-2y โ e-3y), for y > 0,
f (y) = 0, elsewhere.
x1 + x2 = y = 6๐โ3๐ฅ1โ2๐ฅ2
๐ฆโ๐ฅ2
0
๐ฆ
0๐๐ฅ1๐๐ฅ2
0 y
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Example: If X1 and X2 are independent random variables
having U(0,1), find the distribution function of Y = X1+X2.
F(y) = 0, if y โค 0.
๐1 ๐ฅ1 = 1 = ๐2 ๐ฅ2 ๐ ๐ฅ1, ๐ฅ2 ,
for 0 < ๐ฅ1< 1, 0 < ๐ฅ2< 1.
y = x1 + x2
0 1
1
x1
x2
F(y) = 1
2๐ฆ2, if 0 < y โค 1.
F(y) = 1 โ(2โ๐ฆ)2
2, if 1 < y โค 2.
F(y) = 1, if y > 2.
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7.3 Transformation Technique: One Variable
For discrete random variable, whether X and Y = u(X)
is one-to-one or not, finding the distribution of Y is
straight forward substitution.
Example: Let X be the number of heads in tossing a
balanced coin three times, find the probability
distribution of Y = 1/(1+X) . (One-to-one function)
x 0 1 2 3
f(x) 1/8 3/8 3/8 1/8
y 1 1/2 1/3 1/4
g(y) 1/8 3/8 3/8 1/8 12
Example: Let X be the number of heads in tossing a
balanced coin three times, find the probability
distribution of Y = (1 - X)2 . (Not one-to-one function)
x 0 1 2 3
f(x) 1/8 3/8 3/8 1/8
y* 1 0 1 4
g(y*) 1/8 3/8 3/8 1/8
y 0 1 4
g(y) 3/8 4/8 1/8
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Inverse Function Theorem :
For functions of a single variable, if u is a continuously
differentiable function with nonzero derivative at the
point x, then u is invertible in a neighborhood of x, the
inverse is continuously differentiable, and
where y = u(x).
๐ท๐ฆ๐ขโ1(๐ฆ) =1
๐ท๐ฅ๐ข(๐ฅ)
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Theorem 7.1: (Univariate Transformation Theorem)
Let f(x) be the probability density of the continuous
random variable X at x. If the function given by y =
u(x) is differentiable and either (monotone) increasing
and decreasing for all values within the range of X that
has density, then the equation y = u(x) is one-to-one and
x = w(y), and the probability density of Y = u(X) is
given by
g ๐ฆ = ๐ ๐ค ๐ฆ โ ๐คโฒ ๐ฆ , for ๐ขโฒ(๐ฅ) โ 0
0, elsewhere.
(๐ขโ1 ๐ฆ )โฒ = ๐คโฒ(๐ฆ) =1
๐ขโฒ ๐ฅ=
๐๐ฅ
๐๐ฆ
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g ๐ฆ = ๐ ๐ขโ1 ๐ฆ โ
๐
๐๐ฆ๐ขโ1 ๐ฆ , for ๐ขโฒ(๐ฅ) โ 0
0, elsewhere.
Another version of the formula: u-1 (y) = w(y)
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y
x
y = u(x)
b
a
w(a) w(b)
P(u(X) โค y) = P(X โค w(y))
u(x) is increasing function
x
y
Y = 2X
0 .5 1
0 1 2
f(x)
g(y)
Example: X ~ U (0, 1)
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y
x
y = u(x) b
a
w(b) w(a)
P(u(X) โค y) = P(X w(y))
u(x) is decreasing function
x
y
Y = -2X
0 .5 1
- 2 - 1 0
f(x)
g(y)
Example: X ~ U (0, 1)
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Let u(x) be a strictly decreasing function in the range
of X, and w be the inverse function of u, i.e., u-1 .
G(y) = P(Y โค y) = P(u(X) โค y)
= P(X w(y)) = 1 - P(X < w(y)) =1 - F(w(y))
g(y) = G (y) = - F (w(y)) = -f(w(y))wโฒ(y)
Since u(x) is a decreasing function then wโฒ(y) < 0. If
u(x) is a increasing function in the range of X, then
G(y) = P(Y โค y) = P(u(X) โค y)
= P(X โค w(y)) = P(X < w(y)) = F(w(y))
g(y) = G (y) = F (w(y)) = f(w(y))wโฒ(y), with wโฒ(y) > 0.
g ๐ฆ = ๐ ๐ค ๐ฆ โ ๐คโฒ ๐ฆ , for ๐ขโฒ(๐ฅ) โ 0
0, elsewhere.
Function of Random Variables
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Example: Let X have the exponential distribution with
p.d.f. f(x) given by
find the p.d.f. of the random variable Y = ๐. Sol: For y > 0, y = ๐ฅ x = y2 .
w(y) = y2 , w(y) = 2y
g(y) = ๐(๐ฆ2 ) โ 2๐ฆ = ๐โ๐ฆ2|2๐ฆ|
๐ ๐ฅ = ๐โ๐ฅ, for ๐ฅ > 0,0, elsewhere
g ๐ฅ = 2๐ฆ๐โ๐ฆ2, for ๐ฆ > 0,
0, elsewhere. (Weibull Distribution)
for y > 0
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Example: Let X be the a random variable takes the
distance for 0 to a point on the x-axis where the double
arrow will point to, when it is spun. The random
variable Q is the angle that has uniform density
find the p.d.f. of the random variable X.
๐ ๐ = 1
๐, for โ
๐
2< ๐ <
๐
2,
0, elsewhere
a q
0 x
x = a tanq
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a q
0 x
x = a tanq
Sol: x = a tan q ๐๐
๐๐ฅ=
๐
๐2 + ๐ฅ2
1
๐โ
๐
๐2 + ๐ฅ2 g(x) =
=1
๐โ
๐
๐2 + ๐ฅ2 for - < x < .
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Example: If F(x) is the distribution function of the
continuous random variable X, find the p.d.f. of Y = F(x).
Sol: Let y = F(x), )()( xfxFdx
dy
.0)(for ,)(
11 xf
xf
dx
dydy
dx
g(y) = f(x) = 1, for 0 < y <1. )(
1
xf
* Distribution function technique for random number
generation using U(0,1) random number generator.
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Distribution Function Method for Random Numbers:
1. Generate a U(0, 1) random number
2. set this random numbers equal to F(x) and solve
for x.
3. The value x would be a random number from the
distribution that has a distribution function F(x).
Example: Generate a random number from
exponential distribution with parameter q using
U(0,1) random number.
F(x) = 1 โ e-x/q = u u is a random # from U(0, 1)
The random # from the exponential distribution
would be: x = -q ln (1 โ u) 24
Example: If X has the standard normal distribution find
the probability density of Z = X 2.
Sol: z = x 2 is not one-to-one.
First let Y = |X|, then Z = Y 2 = X 2
p.d.f. of Y g(y)= 2n(y; 0, 1) = 2
2๐๐โ
12๐ฆ2
for y > 0
z = y2 , w(z) = y = ๐ง
p.d.f. of Z h(z)= g( ๐ง ) |w(y)|
=2
2๐๐โ
12๐ง
1
2๐งโ
12 =
1
2๐๐งโ
12๐โ
12๐ง
for z > 0
(Chi-square distribution with degrees of freedom = 1.)
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Example: Let X ~ U(0,1), and Y = ๐๐ , find ๐. ๐. ๐. of ๐.
g ๐ฆ = 1
๐๐ฆ
1๐โ1, 0 โค ๐ฆ โค 1
0, elswhere. Answer:
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7.4 Transformation Method: Several Variables
For random variable Y = u(X1, X2) where the joint
distribution or density of X1 and X2 is given, and one
can find the joint distribution or density for Y and X2 or
X1 and Y by holding the other variable fixed, if possible,
and then find the marginal distribution or density
function for Y.
In continuous case, one can first use the transformation
technique with the formula, by holding x1 or x2 fixed,
g ๐ฆ, ๐ฅ2 = ๐(๐ฅ1, ๐ฅ2) โ๐๐ฅ1
๐๐ฆ
g ๐ฅ1, ๐ฆ = ๐(๐ฅ1, ๐ฅ2) โ๐๐ฅ2
๐๐ฆ
or
then find the marginal density of Y.
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Example: If X1 and X2 are independent random having
Poisson distribution with the parameters l1 and l2 , find
the probability distribution of the random variable Y =
X1 + X2.
Sol: Since X1 and X2 are independent, the joint density is
๐ ๐ฅ1, ๐ฅ2 =๐โ๐1๐1
x1
๐ฅ1!โ๐โ๐2๐2
x2
๐ฅ2!=
๐โ(๐1+๐2)๐1x1๐2
x2
๐ฅ1! ๐ฅ2!
for x1 = 0, 1, 2, โฆ, and x2 = 0, 1, 2, โฆ .
Since y = x1 + x2 then x1 = y - x2
๐ ๐ฆ, ๐ฅ2 =๐โ(๐1+๐2)๐1
๐ฆโx2๐2x2
(๐ฆ โ ๐ฅ2)! x2!
for y = 0, 1, 2, โฆ, and x2 = 0, 1, 2, โฆ, y .
Joint Distribution
of Y and X2
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โ(๐ฆ) = ๐โ(๐1+๐2)๐1
๐ฆโx2๐2x2
(๐ฆ โ ๐ฅ2)! x2!
๐ฆ
๐ฅ2=0
for y = 0, 1, 2, โฆ .
=๐โ(๐1+๐2)
๐ฆ!
๐ฆ!
(๐ฆ โ ๐ฅ2)! x2! ๐1
๐ฆโx2๐2x2
๐ฆ
๐ฅ2=0
=๐โ ๐1+๐2 (๐1 + ๐2)๐ฆ
๐ฆ!
The sum of two independent Poisson random
variables with parameters l1 and l2 is a Poisson
random variable with parameter l1 + l2 .
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Example: Let random variables X1 and X2 have the
joint p.d.f. as
find the p.d.f. of the random variable Y =
Sol: Since y decreases as x2 increases and x1 hold
constant, we can find a joint density of X1 and Y and
then use transformation technique to find density of Y.
๐ ๐ฅ1, ๐ฅ2 = ๐โ(๐ฅ1+๐ฅ2), for ๐ฅ1 > 0, ๐ฅ2 > 0,
0, elsewhere
๐1
๐1+๐2
๐ฆ =๐ฅ1
๐ฅ1 + ๐ฅ2โน ๐ฅ2 = ๐ฅ1 โ
1 โ ๐ฆ
๐ฆ for 0 < y < 1
โน ๐๐ฅ2
๐๐ฆ= โ
๐ฅ1
๐ฆ2 ๐ ๐ฅ1, ๐ฆ = ๐โ๐ฅ1/๐ฆ โ๐ฅ1
๐ฆ2 30
๐ ๐ฅ1, ๐ฆ = ๐โ๐ฅ1/๐ฆ โ๐ฅ1
๐ฆ2
=๐ฅ1
๐ฆ2 ๐โ๐ฅ1/๐ฆ for x1 > 0, 0 < y < 1.
โ ๐ฆ = ๐ฅ1
๐ฆ2
โ
0
โ ๐โ๐ฅ1/๐ฆ๐๐ฅ1 Let u = x1 / y
du = -1/y2 dx1
= ๐ขโ
0
โ ๐โ๐ข๐๐ฅ1
= 1 for 0 < y < 1.
It is a U(0, 1)!!!
Function of Random Variables
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Example: Let random variables X and Y have the joint
p.d.f. as
find the joint p.d.f. of X and Z = X + Y & marginal p.d.f.
of Z.
๐ ๐ฅ, ๐ฆ = 2, for ๐ฅ > 0, ๐ฆ > 0, ๐ฅ + ๐ฆ < 10, elsewhere
z = x + y y = z โ x , and 0 < z โ x and 0 < z < 1
x
z
0 1
1
z = x ๐ ๐ฅ, ๐ง = ๐(๐ฅ, ๐ฆ)๐๐ฆ
๐๐ง
= 2 โ 1 = 2
โ ๐ง = 2๐๐ฅ = 2๐ฅ ๐ง0
= 2๐ง๐ง
0
for x < z, 0 < z < 1
for 0 < z < 1 32
Theorem 7.2: (Generalization of Theorem 7.1)
Let f(x1, x2) be the joint probability density of the continuous
random variable X1 and X2. If the function given by y1 = u1(x1,
x2) and y2 = u2(x1, x2) are partially differentiable w.r.t. both x1
and x2 and are one-to-one for which f(x1, x2) โ 0, โ x1, x2 in
the space of X1 and X2 , and the inverse functions x1 = w1(y1,
y2) and x2 = w2(y1, y2) can be uniquely determined (by solving
for x1 and x2), the joint p.d.f. of Y1 = u1(X1, X2) and Y2 = u2(X1,
X2) is
where J is the Jacobian of the transformation, is the
determinant
g ๐ฆ1, ๐ฆ2 = ๐[w1(y1, y2),w2(y1, y2)] โ ๐ฝ
๐ฝ =
๐๐ฅ1
๐๐ฆ1
๐๐ฅ1
๐๐ฆ2
๐๐ฅ2
๐๐ฆ1
๐๐ฅ2
๐๐ฆ2
33
Example: Let random variables X1 and X2 have the
joint p.d.f. as
a) find the joint p.d.f. of Y1= X1 + X2 , and Y2 =
๐ ๐ฅ1, ๐ฅ2 = ๐โ(๐ฅ1+๐ฅ2), for ๐ฅ1 > 0, ๐ฅ2 > 0,
0, elsewhere
๐1
๐1+๐2.
๐ฆ1 = ๐ฅ1 + ๐ฅ2 and ๐ฆ2 =๐ฅ1
๐ฅ1 + ๐ฅ2
โน ๐ฅ1 = ๐ฆ1๐ฆ2 and ๐ฅ2 = ๐ฆ1(1 โ ๐ฆ2)
๐ฝ =
๐๐ฅ1
๐๐ฆ1
๐๐ฅ1
๐๐ฆ2
๐๐ฅ2
๐๐ฆ1
๐๐ฅ2
๐๐ฆ2
=๐ฆ2 ๐ฆ1
1 โ ๐ฆ2 โ๐ฆ1= โ๐ฆ1
0 < y1 and 0 < y2 <1
34
for 0 < y1 and 0 < y2 <1, and 0 elsewhere.
g ๐ฆ1, ๐ฆ2 = ๐โ๐ฆ1 โ๐ฆ1 = ๐ฆ1๐โ๐ฆ1
b) Find the marginal density of Y2 .
โ ๐ฆ2 = ๐ ๐ฆ1, ๐ฆ2 ๐๐ฆ1
โ
0
= ๐ฆ1๐โ๐ฆ1 ๐๐ฆ1
โ
0
= (2)
= 1
for 0 < y2 <1, and 0 elsewhere.
35
Example: Let random variables X1 and X2 have the
joint p.d.f. as
a) find the joint p.d.f. of Y = X1 + X2 , and Z = X2 .
๐ ๐ฅ1, ๐ฅ2 = 1, for 0 < ๐ฅ1 < 1,0 < ๐ฅ2 < 1,0, elsewhere
๐ฝ =
๐๐ฅ1
๐๐ฆ
๐๐ฅ1
๐๐ง๐๐ฅ2
๐๐ฆ
๐๐ฅ2
๐๐ง
=1 โ10 1
= 1
y = x1 + x2 , and z = x2
x1= y - x2 , and x2 = z, z < y < z + 1, 0 < z < 1
y
z
0 1 2
1
y = z + x1 , z = x2
36
๐ ๐ฆ, ๐ง = 1 โ 1 = 1,
for z < y < z + 1, 0 < z < 1, and 0, else where.
b) Find the marginal density of Y.
โ ๐ฆ =
0, ๐ฆ โค 0
1๐๐ง = ๐ฆ,๐ฆ
0
for 0 < ๐ฆ < 1
1๐๐ง = 2 โ ๐ฆ, 1
๐ฆโ1
for 1 < ๐ฆ < 2
0, ๐ฆ โฅ 2
y
z
0 1 2
1
Function of Random Variables
FRV - 7
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Example: Let random variables X1 and X2 have the
joint p.d.f. as
a) find the joint p.d.f. of Z = X + Y , and W = X - Y.
๐ ๐ฅ, ๐ฆ = 2, for ๐ฅ > 0, ๐ฆ > 0, ๐ฅ + ๐ฆ < 1,0, elsewhere
๐ฝ =
1
2
1
21
2โ
1
2
= โ1/2
z = x + y and w = x - y x=(z + w)/2; y=(z-w)/2,
z + w > 0, z - w > 0, and 0< z < 1.
x
z
0 1
1
-1
z โ w = 0
z + w = 0
๐ ๐ง,๐ค = 2 โ โ1
2= 1
for 0 < z < 1, z > -w, and z > w. 38
b) find the marginal p.d.f. of Z = X + Y .
z
w
0 1
1
-1
z โ w = 0
z + w = 0
๐ ๐ง = ๐(๐ง, ๐ค)๐ง
โ๐ง
๐๐ค
for 0 < z < 1.
= 1๐ง
โ๐ง
๐๐ค = ๐ค ๐ง
โ๐ง= 2๐ง
๐ ๐ง, ๐ค = 1, for 0 < z < 1, z > -w, and z > w.
39
Example: Let random variables X1 and X2 have the
joint p.d.f. as
find the marginal p.d.f. of Z = X + Y .
๐ ๐ฅ, ๐ฆ = 2, for ๐ฅ > 0, ๐ฆ > 0, ๐ฅ + ๐ฆ < 1,0, elsewhere
๐ ๐ง = ๐(๐ฅ, ๐ง)๐ง
0
๐๐ฅ
for 0 < z < 1.
= 2๐ง
0
๐๐ฅ = 2๐ฅ ๐ง0
= 2๐ง
Previous Example: g ๐ฅ1, ๐ฆ = ๐(๐ฅ1, ๐ฅ2) โ๐๐ฅ2
๐๐ฆ
z = x + y y = z โ x
๐ ๐ฅ, ๐ง = ๐(๐ฅ, ๐ฆ) โ๐๐ฆ
๐๐ง = 2โ 1 = 2
for 0 < z < 1 and 0 < z โ x .
x
z
0 1
1 z โ x = 0
40
Example: Let random variables X1 and X2 have the
joint p.d.f. as
a) find the joint p.d.f. of Y1 = X12 and Y2 = X1 X2 .
๐ ๐ฅ1, ๐ฅ2 = 4๐ฅ1๐ฅ2 , for 0 < ๐ฅ1 < 1,0 < ๐ฅ2 < 1
0, elsewhere
๐ฝ =
1
2 ๐ฆ10
โ1
2๐ฆ2๐ฆ1
โ3/21
๐ฆ1
=1
2๐ฆ1
y1 = x12 and y2 = x1x2 x1= ๐ฆ1 , x2 =
๐ฆ2
๐ฆ1,0 <y2< ๐ฆ1
๐ ๐ฆ1, ๐ฆ2 = 4 ๐ฆ1 โ๐ฆ2
๐ฆ1๐ฝ
= 4y2 /2y1 =2y2 /y1
y1
y2
0 1
1 y2
2 = y1
for 0 < y2< ๐ฆ1
y22 < y1
41
X1, X2 , โฆ , Xn ~ f(x1, x2 , โฆ , xn)
Y1 ~ u1(x1, x2 , โฆ , xn)
Y2 ~ u2(x1, x2 , โฆ , xn)
Yn ~ un(x1, x2 , โฆ , xn) โฆ
g(x1, x2 , โฆ , xn) = f(x1, x2 , โฆ , xn)|J|
๐ฝ =
๐๐ฅ1
๐๐ฆ1โฆ
๐๐ฅ1
๐๐ฆ๐
โฎ โฑ โฎ๐๐ฅ๐
๐๐ฆ1โฏ
๐๐ฅ๐
๐๐ฆ๐
42
7.5 Moment-Generating Function Technique
Theorem 7.3 (Generalized Version): If X1, X2, โฆ, Xn
are independent random variables with m.g.f.โs is
i = 1, 2,โฆ, n, then the m.g.f. of is
)(tMiX
n
i
iXY taMtMi
1
)()(
n
i ii XaY1
Function of Random Variables
FRV - 8
43
Example: Find the probability distribution of the
sum of n independent random variables X1, X2, โฆ, Xn
that have Poisson distribution with parameters l1, l2,
โฆ, ln , respectively.
)1()(
-
ti
i
e
X etMl
The m.g.f. of Poisson distribution is
So, for Y = X1+ X2 + โฆ+ Xn , the m.g.f. is
)1)((
1
)1( 21)(-+++
-
tn
ti e
n
i
e
Y eetMllll
which is the m.g.f. of a Poisson distribution with
parameter l = l1+ l2 + โฆ+ ln , therefore, Y has a
Poisson distribution with l = l1+ l2 + โฆ+ ln . 44
Example: Find the probability distribution of the
sum of n independent random variables X1, X2, โฆ, Xn
that have Poisson distribution with parameters l1, l2,
โฆ, ln , respectively.
)1()(
-
ti
i
e
X etMl
The m.g.f. of Poisson distribution is
So, for Y = X1+ X2 + โฆ+ Xn , the m.g.f. is
)1)((
1
)1( 21)(-+++
-
tn
ti e
n
i
e
Y eetMllll
which is the m.g.f. of a Poisson distribution with
parameter l = l1+ l2 + โฆ+ ln , therefore, Y has a
Poisson distribution with l = l1+ l2 + โฆ+ ln .
45
Example: If X1, X2, โฆ, Xn are mutually independent
random variables from normal distributions with
means m1, m2, m3, โฆ, mn, and variances s12, s2
2, s32,
โฆ, sn2, then the linear function
has the normal distribution N(Scimi , Sci2si
2).
n
i
ii XcY1
.1
if :Mean Samplen
cXY i
46
Sol:
+
+
2
1 1
2/
2
1
22
1
222
)()(
tctc
n
i
n
i
tctc
iXY
n
iii
n
i
ii
iiii
i
e
etcMtM
sm
sm
m.g.f. of
n
i
n
i
ii iiccN
1
22
1
, sm
47
Distribution of
If X1, X2, โฆ, Xn are observations of a random
sample of size n from the normal distribution
N(m, s 2), then the distribution of the sample
mean is N(m, s 2/n)
n
i
iXn
X1
1
X
nX
X
ss
mm