Introduction to Probability & Statistics Joint Expectations

Properties of Expectations Recall that: 1. E[c] = c 2. E[aX + b] = aE[X] + b 3. 2(ax + b) = a22 4. E[g(x)] = g x dF ( ) 

Joint Expectation Let Z = X + Y E[Z] = E[X] + E[Y]  ( ) cov( , z x y 2 ( ) cov( , z x y  

Derivation   E Z zf x y dxdy [ ] ( , )    ( ) , x y f dxdy Let Z = X + Y E Z zf x y dxdy XY [ ] ( , )      ( ) , x y f dxdy XY

Derivation E[Z]      E Z zf x y dxdy [ ] ( , )    ( ) , x y f Let Z = X + Y E Z zf x y dxdy XY [ ] ( , )      ( ) , x y f dxdy XY    y XY x xf dxdy yf ( , )    x f y dydx dxdy XY ( , )    xf x dx yf y dy X Y ( ) = E[X] + E[Y]

Derivation of 2(z)   ( ) , [ ] z f x y dxdy E Z   Miracle 13 occurs = 2(x) + 2(y)

In General In general, for Z = the sum of n independent variates, Z = X1 + X2 + X3 + . . . + Xn E Z X i n [ ]   1  2 ( ) z x

Class Problem Suppose n customers enter a store. The ith customer spends some Xi amount. Past data indicates that Xi is a uniform random variable with mean $100 and standard deviation $5. Determine the mean and variance for the total revenue for 5 customers.

Class Problem Total Revenue is given by TR = X1 + X2 + X3 + X4 + X5 Using the property E[Z] = E[X] + E[Y], E[TR] = E[X1] + E[X2] + E[X3] + E[X4] + E[X5] = 100 + 100 + 100 + 100 + 100 = $500

Class Problem TR = X1 + X2 + X3 + X4 + X5 Use the property s2(Z) = s2(X) + s2(Y). Assuming Xi, i = 1, 2, 3, 4, 5 all independent s2(TR) = s2(X1) + s2(X2) + s2(X3) + s2(X4) + s2(X5) = 52 + 52 + 52 + 52 + 52 = 125

Class Problem TR = X1 + X2 + X3 + X4 + X5 Xi , independent with mean 100 and standard deviation 5 E[TR] = $500 s2(TR) = 125