Chapter 6 - Lecture notes 6 PDF

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Lucas Godoy...

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6.3 The Method of Distribution Functions Illustrate method of distribution functions with a univariate example: • If Y has probability density function f(y) and if U is some function of Y then we can find F (u) = P(U 0 is a constant. a. Derive the density function U = cY b. Identify the density of U as one types we studied in chapter 4. Be sure to identify any parameters values c. The parameters α and β of a gamma distributed random variable are “shape” and “scale” parameters. How do the scale and shape parameters for U compare to those for Y?

a. 1. Distribution function of U is 2. Replace U by cY on right hand side 3. We can contain density function by taking derivative e of F (u) with respect to u

b. U ~ Gamma(α, cβ) c. The shape parameters coincide. On the other hand, the scale parameter for U is c times the scale parameter for Y. 6.4 Method of Transformation • Random variable Y • Let U = h(Y) where h(y) is either increasing or decreasing • To use this method, we do not need to know the CDF function of the Random Variable Y knowing pdf is enough then multiply by dh / du In Example above we worked with a random variable Y (amount of sugar produced) with a density function given by we were interested in a new random variable (profit) given by U = 3Y - 1. Find probability density function for U by transformation method

Let Y have the pdf given by

find the density of U = -4Y + 3

6.5 Method of Moment Generating Functions • Theorem 6.1 (Uniqueness Theorem): let mₓ(t) and m (t) and denote the moment-generating functions of random variables X and Y, respectively. If both moment-generating functions exist and mₓ(t) = m (t) for all values t, then X and Y have the same probability distribution ◦ m (t) = E(e ) Suppose Y ~ N( , ) show that has normal distribution, a normal distribution with mean 0 and variance 1 Solution We have seen that Y - µ has moment generating function e Hence

Let Z be normally distributed random variable with mean 0 and variance 1. Use method of moment generating functions to find probability distribution of of Z²

• Method of moment generating functions is often useful for finding distribution of sums of independent random variables Theorem 6.2: Let Y , Y , ..., Y be independent random variables with moment generating functions m (t), m (t), ..., m (t) respectively. If U = Y₁ + Y₂ + ... + Y then m (t) = m (t) × m (t) × ... × m (t) The number of customer arrivals at a checkout counter in a given interval of time possesses approx a Poisson probability distribution. If Y₁ denotes the time until the first arrival Y₂ denotes the time between the first and second arrival, ..., and Y denoted the time between the (n-1)st and nth arrival, then it can be shown that Y₁, Y₂, ..., Y are independent random variables, with density function for Y given by Find the probability density function for the waiting time from the opening of the counter until the nth customer arrives (We want density function U = Y₁ + Y₂ + ... + Y )

Theorem 6.3 Let Y , Y , ..., Y be independent normally distributed random variables with E(Y ) = µ and V(Y ) = σ², for i = 1, 2, ..., n, and let a₁, a₂, ..., a be constants if U = a₁Y₁ + a₂Y₂ + a Y . Then U is normally distributed with E(U) = and V(U) = Theorem 6.4 Let Y , Y , ..., Y be independent normally distributed random variables with E(Y ) = µ and V(Y ) = σ² for i = 1, 2, ..., n and define Z = ———, i = 1, ..., n then has X² distribution with n degrees of freedom 6.7 Order Statistics Observe random variables based on their magnitudes • Let Y₁, Y₂,..., Y denote independent continuous random variables with distribution function F(y) and density function f(y). We denote ordered random variables Y by Y ₁ , Y ₂ , ..., Y where Y ₁ < Y ₂ < ... < Y Using this is the max value of random variables • The density functions for Y ₁ and Y can be found using method of distribution functions. Let’s note cdf for Y is ◦ So the density function for Y is • Pdf for Y is • pdf derived......