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§3.5 Distribution of Special Functions. Sum and difference Z=X+Y, Z=X-Y Product and quotient Z=XY, Z=YX. Functions of discrete random vectors. Suppose that (X, Y) ~ P(X = x i , Y = y j ) = p ij , i, j = 1, 2, …
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§3.5 Distribution of Special Functions • Sum and difference Z=X+Y, Z=X-YProduct and quotientZ=XY, Z=Y\X
Functions of discrete random vectors Suppose that (X, Y)~P(X=xi, Y=yj)=pij ,i, j=1, 2, … then Z=g(X, Y)~P{Z=zk}= =pk , k=1, 2, … or
EX Suppose that X and Y are independent, and the pmfs of X and Y are Find the pmf of Z=X+Y. Solution Because X and Y are independent, so
EXSuppose that X and Y are independent and both are uniformly distributed on 0-1 with law X01 • P q p • Try to determine the distribution law of • W=X+Y ;(2) V=max(X, Y); • (3) U=min(X, Y); • (4)The joint distribution law of w and V .
2 0 1 1 0 1 1 1 0 0 0 1 W 0 1 2 V 0 1 0 0 0
Example Suppose and are independent of each other, then ExampleSuppose and are independent of each other , then
Example Suppose and are independent of each other, then Proof i = 0 , 1 , 2 , … j = 0 , 1 , 2 , … then
r = 0 , 1 , … then
a) Sum and difference Let X and Y be continuous r.v.s with joint pdf f (x, y), Let Z=X+Y ,Find the pdf of Z=X+Y. The cdf of Z is : D={(x, y): x+y ≤z}
In particular, if X and Y are independent, then which is called the convolution of fx(.) and fy(.).
Example 1 If X and Y are independent with the same pdf Find the pdf of Z=X+Y. Solution i.e.
So when or , when, when , so
Example 2 If X and Y are independent with the same distributionN(0,1) , Find the pdf of Z=X+Y. Solution
let then Z=X+Y ~N(0,2).
let f(x,y) is the joint pdf of (X,Y), then the pdf of Z=Y\X is When X and Y are independent, 二、Z=Y\X, Z=XY