{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Example_class_7_handout_solution

Example_class_7_handout_solution - THE UNIVERSITY OF HONG...

This preview shows pages 1–4. Sign up to view the full content.

THE UNIVERSITY OF HONG KONG DEPARTMENT OF STATISTICS AND ACTUARIAL SCIENCE STAT1301 PROBABILITY AND STATISTICS I EXAMPLE CLASS 7 Review Quantiles of a distribution The β quantile of a probability distribution function F X of a random variable X is defined to be F - 1 X ( β ) = inf { x R : F X ( x ) β } . Transformation of pdf Let X be a continuous random variable distributed on a space S with pdf f x ( x ). Let Y = g ( X ) where g is a function such that g - 1 exists. Then the pdf of Y can be obtained by f Y ( y ) = f X ( g - 1 ( y )) d dy g - 1 ( y ) , y g ( S ) . Problems Problem 1. Let X Γ( α, λ ) , where α, λ > 0 . The pdf of X is f ( x ) = λ α Γ( α ) x α - 1 e - λx if x > 0 , 0 if x 0 . Find the mean and variance of X . Solution. 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
E ( X ) = ˆ 0 xf ( x ) dx = ˆ 0 x · λ α Γ( α ) x α - 1 e - λx dx = λ α Γ( α ) ˆ 0 x α e - λx dx = λ α Γ( α ) · Γ( α + 1) λ α +1 = α λ E ( X 2 ) = ˆ 0 x 2 f ( x ) dx = ˆ 0 λ α Γ( α ) x α +1 e - λx dx = λ α Γ( α ) ˆ 0 x α +1 e - λx dx = λ α Γ( α ) · Γ( α + 2) λ α +2 = α ( α + 1) λ 2 V ar ( X ) = E ( X 2 ) - E ( X ) 2 = α ( α + 1) λ 2 - α 2 λ 2 = α λ 2 Problem 2. The random variable X is said to have a normal distribution with mean μ and variance σ 2 (Gaussian distribution) if its pdf is defined by f ( x ) = 1 2 πσ e - ( x - μ ) 2 2 σ 2 , x ( -∞ , ) . (a) Show that ´ -∞ e - x 2 / 2 dx = 2 π . Hence show that ´ -∞ f ( x ) dx = 1. (b) Find the moment generating function of X. Hence show that the mean is equal to μ and the variance is equal to σ 2 . 2
(c) Let Z = X - μ σ . Show that Z is normal with mean 0 and variance 1. (d) Show that E ( Z k ) = 0 if k is an odd integer. (e) Find E ( | Z | ). (f) Find the mgf of Z 2 . What is the distribution of Z 2 ? Solution. (a) Note that ˆ -∞ e - x 2 / 2 dx 2 = ˆ -∞ ˆ -∞ e - x 2 + y 2 2 dxdy Use the change of variables x = r cos θ , y = r sin θ . Then ˆ -∞ ˆ -∞ e - x 2 + y 2 2 dxdy = ˆ 2 π 0 ˆ 0 e - ( r 2 cos 2 θ + r 2 sin 2 θ ) / 2 ( x, y ) ( r, θ ) drdθ where ( x,y ) ( r,θ )

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 12

Example_class_7_handout_solution - THE UNIVERSITY OF HONG...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online