Solution:
The formula for the expected loss is E[L(X,theta)] = k*sigma^2 + k*(mu-theta)^2 when L(X,theta) = k*(X-theta)^2 is the quality loss function.
Applied to this problem, the target theta is 12, the mean is 11, and the standard deviation sigma is 2. So the expected loss is 10*(2^2 + (11-12)^2) = 100.
Solution:
Recall the series
1 1 ----- = 1 + z + z^2 + ... => ----- = 1 + 1/z + 1/z^2 + ... 1 - z 1-1/zwhich leads to the z-transform of the signal 1,1,1,1,... To insert one zero between every pair of ones, we substitute z by z^2, and we obtain z^2/(z^2 - 1) as the z-transform of the signal 1,0,1,0,... After multiplication with 2/z, we obtain the z-transform of the signal 0,2,0,2,... The addition of these two z-transforms
z^2 2 z^2 z^2 + 2 z ------- + z^(-1) ------- = ----------- z^2 - 1 z^2 - 1 z^2 - 1yields the z-transform of the signal 1,2,1,2,...
Solution:
If we know the output of the filter when giving it the unit impulse on input, then we can determine the output for any input. In formulas, let h_k, k=0,1,.., infinity, be the inpulse response, then
infinity -------- \ -k H(z) = > h z / k -------- k = 0is the transfer function of the filter. In the time domain, for periodic input u, we have that the output y is y = h*u, with * the circular convolution operator.
minimize x1 + 2 x2 subject to { 6 x1 + x_2 ≥ 6 { x1 + x_2 ≥ 3 { 2 x1 + 5 x_2 ≥ 10
Solution:
Graphing the three constraints and drawing rays parallel to the objective function x1 + 2*x2, we see that the minimum occurs at the intersection of the second and third line. Solving this linear system, we obtain (x1,x2) = (5/3,4/3) with value 13/3 of the objective function.
As an alternative to the graphical solution, we could solve three linear systems and compare the values of the intersection points at the objective.
Another method (recommended for larger problems) is to set up the tableau and execute the simplex algorithm.
Solution:
We apply the inverse continuous compounding rule:
-0.05 x 5 $1,000 e = $778.80.
Solution: The formulas for elasticity e and revenue R are
D'(p) e = - ----- and R = p D(p) D(p)where p is the unit price and D(p) the demand function. A simple derivation of revenue with respect to price
dR ( D'(p)) -- = D(p) + p D'(p) = D(p)( 1+p -----) = D(p)(1-e) dp ( D(p) )shows the relation between change in revenue due to price increase and the elasticity of demand.
The importance of this relation follows. If e < 1, then 1-e > 0 and the revenue will go up when the price increases. If e > 1, then 1-e < 0 and the revenue will go down when the price increases. So elasticity determines whether the producer can increase the price and increase the revenue at the same time.
Solution:
There are several good answers to this question, here are a few suggestions:
Expanding a bit on these suggestions makes a good answer.