Download TM 663 Operations Planning

Introduction to
Probability & Statistics
Joint Expectations
Properties of Expectations
Recall that:
1.
2.
3.
4.
E[c] = c
E[aX + b] = aE[X] + b
2(ax + b) = a22
E[g(x)] =  g( x)dF ( x)
Joint Expectation
Let Z = X + Y
E[Z] = E[X] + E[Y]
 ( z)   ( x)   ( y)  2 cov( x, y)
2
2
2
Derivation
Let Z = X + Y
E [ Z ]  xyzfXY ( x, y) dxdy
 xy( x  y) f XY ( x, y) dxdy
Derivation E[Z]
Let Z = X + Y
E [ Z ]  xyzfXY ( x, y) dxdy
 xy( x  y) f XY ( x, y) dxdy
 xy xfXY ( x, y) dxdy  xyyfXY ( x, y) dxdy
 xxy f XY ( x, y) dydx  yyx f XY ( x, y) dxdy
 xxfX ( x) dx  yyfY ( y) dy
= E[X] + E[Y]
Derivation of 2(z)
2
2
2
 ( z)  xyz f XY ( 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
n
E [ Z ]   E [ Xi ]
i 1
n
2
 ( z)    ( xi )
2
i 1
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 2(Z) = 2(X) + 2(Y).
Assuming Xi, i = 1, 2, 3, 4, 5 all independent
2(TR) = 2(X1) + 2(X2) + 2(X3) + 2(X4) + 2(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
2(TR) = 125