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Functions of Random Variables. Methods for determining the distribution of functions of Random Variables. Distribution function method Moment generating function method Transformation method. Distribution function method. Let X, Y, Z …. have joint density f ( x,y,z, … )
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Methods for determining the distribution of functions of Random Variables • Distribution function method • Moment generating function method • Transformation method
Distribution function method Let X, Y, Z …. have joint density f(x,y,z, …) Let W = h( X, Y, Z, …) First step Find the distribution function of W G(w) = P[W ≤ w] = P[h( X, Y, Z, …)≤ w] Second step Find the density function of W g(w) = G'(w).
Example: Student’s t distribution Let Z and U be two independent random variables with: • Z having a Standard Normal distribution and • U having a c2 distribution with n degrees of freedom Find the distribution of
The density of Z is: The density of U is:
Therefore the joint density of Z and U is: The distribution function of T is:
Then where
Student’s t distribution where
Student – W.W. Gosset Worked for a distillery Not allowed to publish Published under the pseudonym “Student
t distribution standard normal distribution
Let x1, x2, … , xndenote a sample of size n from the density f(x). Let M = max(xi) then determine the distribution of M. Repeat this computation for m = min(xi) Assume that the density is the uniform density from 0 to q.
Hence and the distribution function
Differentiating we find the density function of M. f(x) g(t)
Differentiating we find the density function of m. f(x) g(t)
The probability integral transformation This transformation allows one to convert observations that come from a uniform distribution from 0 to 1 to observations that come from an arbitrary distribution. Let U denote an observation having a uniform distribution from 0 to 1.
Let f(x) denote an arbitrary density function and F(x) its corresponding cumulative distribution function. Find the distribution of X. Let Hence.
Thus if U has a uniform distribution from 0 to 1. Then has density f(x). U
Definition Let X denote a random variable with probability density function f(x) if continuous (probability mass function p(x) if discrete) Then mX(t) = the moment generating function of X
The distribution of a random variable X is described by either • The density function f(x) if X continuous (probability mass function p(x) if X discrete), or • The cumulative distribution function F(x), or • The moment generating function mX(t)
Properties • mX(0) = 1
Let X be a random variable with moment generating function mX(t). Let Y = bX + a Then mY(t) = mbX + a(t) = E(e [bX + a]t) = eatmX (bt) • Let X and Y be two independent random variables with moment generating function mX(t) and mY(t) . Then mX+Y(t) = mX (t) mY (t)
Let X and Y be two random variables with moment generating function mX(t) and mY(t) and two distribution functions FX(x) and FY(y) respectively. Let mX (t) = mY (t) then FX(x) = FY(x). This ensures that the distribution of a random variable can be identified by its moment generating function
using or
Moment generating function of the Standard Normal distribution where thus
Note: Also
Note: Also
Using of moment generating functions to find the distribution of functions of Random Variables
Example Suppose that X has a normal distribution with mean mand standard deviation s. Find the distribution of Y = aX + b Solution: = the moment generating function of the normal distribution with mean am + b and variance a2s2.
Thus Y = aX + b has a normal distribution with mean am + b and variance a2s2. Special Case: the z transformation Thus Z has a standard normal distribution .
Example Suppose that X and Y are independent eachhaving a normal distribution with means mX and mY , standard deviations sX and sY Find the distribution of S = X + Y Solution: Now
or = the moment generating function of the normal distribution with mean mX + mY and variance Thus Y = X + Y has a normal distribution with mean mX + mY and variance
Example Suppose that X and Y are independent eachhaving a normal distribution with means mX and mY , standard deviations sX and sY Find the distribution of L = aX + bY Solution: Now
or = the moment generating function of the normal distribution with mean amX + bmY and variance Thus Y = aX + bY has a normal distribution with mean amX + bmY and variance
a = +1 and b = -1. Special Case: Thus Y = X - Y has a normal distribution with mean mX - mY and variance
Example (Extension to n independent RV’s) Suppose that X1, X2, …, Xn are independent eachhaving a normal distribution with means mi, standard deviations si (for i = 1, 2, … , n) Find the distribution of L = a1X1 + a1X2 + …+ anXn Solution: (for i = 1, 2, … , n) Now
or = the moment generating function of the normal distribution with mean and variance Thus Y = a1X1 + … + anXnhas a normal distribution with mean a1m1+ …+ anmn and variance
Special case: In this case X1, X2, …, Xn is a sample from a normal distribution with mean m, and standard deviations s, and
Thus has a normal distribution with mean and variance
Summary If x1, x2, …, xn is a sample from a normal distribution with mean m, and standard deviations s, then has a normal distribution with mean and variance
Sampling distribution of Population