# Week 91 Example A device containing two key components fails when and only when both components fail. The lifetime, T 1 and T 2, of these components are.

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week 91 Example A device containing two key components fails when and only when both components fail. The lifetime, T 1 and T 2, of these components are independent with a common density function given by The cost, X, of operating the device until failure is 2T 1 + T 2. Find the density function of X.

week 92 Convolution Suppose X, Y jointly distributed random variables. We want to find the probability / density function of Z=X+Y. Discrete case X, Y have joint probability function p X,Y (x,y). Z = z whenever X = x and Y = z – x. So the probability that Z = z is the sum over all x of these joint probabilities. That is If X, Y independent then This is known as the convolution of p X (x) and p Y (y).

week 93 Example Suppose X~ Poisson(λ 1 ) independent of Y~ Poisson(λ 2 ). Find the distribution of X+Y.

4 Convolution - Continuous case Suppose X, Y random variables with joint density function f X,Y (x,y). We want to find the density function of Z=X+Y. Can find distribution function of Z and differentiate. How? The Cdf of Z can be found as follows: If is continuous at z then the density function of Z is given by If X, Y independent then This is known as the convolution of f X (x) and f Y (y).

week 95 Example X, Y independent each having Exponential distribution with mean 1/λ. Find the density for W=X+Y.

week 96 Some Recalls on Normal Distribution If Z ~ N(0,1) the density of Z is If X = σZ + μ then X ~ N(μ, σ 2 ) and the density of X is If X ~ N(μ, σ 2 ) then

week 97 More on Normal Distribution If X, Y independent standard normal random variables, find the density of W=X+Y.

week 98 In general, If X 1, X 2,…, X n i.i.d N(0,1) then X 1 + X 2 +…+ X n ~ N(0,n). If,,…, then If X 1, X 2,…, X n i.i.d N(μ, σ 2 ) then S n = X 1 + X 2 +…+ X n ~ N(nμ, nσ 2 ) and

9 Sum of Independent χ 2 (1) random variables Recall: The Chi-Square density with 1 degree of freedom is the Gamma(½, ½) density. If X 1, X 2 i.i.d with distribution χ 2 (1). Find the density of Y = X 1 + X 2. In general, if X 1, X 2,…, X n ~ χ 2 (1) independent then X 1 + X 2 +…+ X n ~ χ 2 (n) = Gamma(n/2, ½). Recall: The Chi-Square density with parameter n is

week 910 Cauchy Distribution The standard Cauchy distribution can be expressed as the ration of two Standard Normal random variables. Suppose X, Y are independent Standard Normal random variables. Let. Want to find the density of Z.

week 911 Change-of-Variables for Double Integrals Consider the transformation, u = f(x,y), v = g(x,y) and suppose we are interested in evaluating. Why change variables? In calculus: - to simplify the integrand. - to simplify the region of integration. In probability, want the density of a new random variable which is a function of other random variables. Example: Suppose we are interested in finding. Further, suppose T is a transformation with T(x,y) = (f(x,y),g(x,y)) = (u,v). Then, Question: how to get f U,V (u,v) from f X,Y (x,y) ? In order to derive the change-of-variable formula for double integral, we need the formula which describe how areas are related under the transformation T: R 2  R 2 defined by u = f(x,y), v = g(x,y).

week 912 Jacobian Definition: The Jacobian Matrix of the transformation T is given by The Jacobian of a transformation T is the determinant of the Jacobian matrix. In words: the Jacobian of a transformation T describes the extent to which T increases or decreases area.

week 913 Change-of-Variable Theorem in 2-dimentions Let x = f(u,v) and y = g(u,v) be a 1-1 mapping of the region Auv onto Axy with f, g having continuous partials derivatives and det(J(u,v)) ≠ 0 on Auv. If F(x,y) is continuous on Axy then where

week 914 Example Evaluate where Axy is bounded by y = x, y = ex, xy = 2 and xy = 3.

week 915 Change-of-Variable for Joint Distributions Theorem Let X and Y be jointly continuous random variables with joint density function f X,Y (x,y) and let D XY = {(x,y): f X,Y (x,y) >0}. If the mapping T given by T(x,y) = (u(x,y),v(x,y)) maps D XY onto D UV. Then U, V are jointly continuous random variable with joint density function given by where J(u,v) is the Jacobian of T -1 given by assuming derivatives exists and are continuous at all points in D UV.

week 916 Example Let X, Y have joint density function given by Find the density function of

week 917 Example Show that the integral over the Standard Normal distribution is 1.

week 918 Density of Quotient Suppose X, Y are independent continuous random variables and we are interested in the density of Can define the following transformation. The inverse transformation is x = w, y = wz. The Jacobian of the inverse transformation is given by Apply 2-D change-of-variable theorem for densities to get The density for Z is then given by

week 919 Example Suppose X, Y are independent N(0,1). The density of is

week 920 Example – F distribution Suppose X ~ χ 2 (n) independent of Y ~ χ 2 (m). Find the density of This is the Density for a random variable with an F-distribution with parameters n and m (often called degrees of freedom). Z ~ F(n,m).

week 921 Example – t distribution Suppose Z ~ N(0,1) independent of X ~ χ 2 (n). Find the density of This is the Density for a random variable with a t-distribution with parameter n (often called degrees of freedom). T ~ t (n)

week 922 Some Recalls on Beta Distribution If X has Beta(α,β) distribution where α > 0 and β > 0 are positive parameters the density function of X is If α = β = 1, then X ~ Uniform(0,1). If α = β = ½, then the density of X is Depending on the values of α and β, density can look like: If X ~ Beta(α,β) then and

week 923 Derivation of Beta Distribution Let X 1, X 2 be independent χ 2 (1) random variables. We want the density of Can define the following transformation

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