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Example

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

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Example

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  1. Example • A device containing two key components fails when and only when both components fail. The lifetime, T1 and T2, of these components are independent with a common density function given by • The cost, X, of operating the device until failure is 2T1 + T2. Find the density function of X. week 10

  2. 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 pX,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 pX(x) and pY(y). week 10

  3. Example • Suppose X~ Poisson(λ1) independent of Y~ Poisson(λ2). Find the distribution of X+Y. week 10

  4. Convolution - Continuous case • Suppose X, Y random variables with joint density function fX,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 fX(x) and fY(y). week 10

  5. Example • X, Y independent each having Exponential distribution with mean 1/λ. Find the density for W=X+Y. week 10

  6. Some Recalls on Normal Distribution • If Z ~ N(0,1) the density of Z is • If X = σZ + μ then X ~ N(μ, σ2) and the density of X is • If X ~ N(μ, σ2) then week 10

  7. More on Normal Distribution • If X, Y independent standard normal random variables, find the density of W=X+Y. week 10

  8. In general, • If X1, X2,…, Xni.i.d N(0,1)then X1+ X2+…+ Xn ~ N(0,n). • If , ,…, then • If X1, X2,…, Xni.i.d N(μ, σ2)then Sn = X1+ X2+…+ Xn ~ N(nμ, nσ2) and week 10

  9. Sum of Independent χ2(1) random variables • Recall: The Chi-Square density with 1 degree of freedom is the Gamma(½ , ½) density. • If X1, X2 i.i.d with distribution χ2(1). Find the density of Y = X1+ X2. • In general, if X1, X2,…, Xn ~ χ2(1) independent then X1+ X2+…+ Xn ~ χ2(n) = Gamma(n/2, ½). • Recall: The Chi-Square density with parameter n is week 10

  10. 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 10

  11. 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 fU,V(u,v) from fX,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: R2 R2 defined by u = f(x,y), v = g(x,y). week 10

  12. 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 10

  13. 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 10

  14. Example • Evaluate where Axy is bounded by y = x, y = ex, xy = 2 and xy = 3. week 10

  15. Change-of-Variable for Joint Distributions • Theorem Let X and Y be jointly continuous random variables with joint density function fX,Y(x,y) and let DXY = {(x,y): fX,Y(x,y) >0}. If the mapping T given by T(x,y) = (u(x,y),v(x,y)) maps DXY onto DUV. 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 DUV . week 10

  16. Example • Let X, Y have joint density function given by Find the density function of week 10

  17. Example • Show that the integral over the Standard Normal distribution is 1. week 10

  18. 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 10

  19. Example • Suppose X, Y are independent N(0,1). The density of is week 10

  20. Example – F distribution • Suppose X ~ χ2(n) independent of Y ~ χ2(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 10

  21. Example – t distribution • Suppose Z ~ N(0,1) independent of X ~ χ2(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 10

  22. Some Recalls on Beta Distribution • If X has Beta(α,β)distribution whereα > 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 10

  23. Derivation of Beta Distribution • Let X1, X2 be independent χ2(1) random variables. We want the density of • Can define the following transformation week 10

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