This is going to be a bit lengthy, but I am stuck and need a bit of guidance. So, Let me start this out by saying I am stuck trying to find the variance of x and y (var(x+y), in the equation 
. Now as in the picture, all values are given to us, but i am trying to understand where each variable is coming from. The book tells me to reference an equation in a different part of the book, which gives me the information, "we have already computed Var(s) = Var(x+y)," which references the table "Calculation of the expected value and variance for total daily sales at DiCarlo Motors", which is this table:
Now here's where the problem lies. The data given to me for the problem is the following table, which is separate from the previous table. This is a bivariate probability distribution for x and y, for clarity.
In the scenario, we have found the expected values of investment return on x and y:
E(x) = compute product for all scenarios for sum(f(x,y)*f(x)) = .10(-40) + .25(5) + .5(15) + .15(30) = 9.25 and
E(y) = compute product for all scenarios for sum(f(x,y)*f(y)) = .10(30) + .25(5) + .5(4) + .15(2) = 6.55.
But, the scenario further expands to find the risk associated with investments x and y giving the equations:
Var(x) = compute product for all scenarios for sum(f(x,y)*(f(x)-E(x)2) = .1(-40-9.25)2 + .25(5-9.25)2 + .50(15-9.25)2+ .15(30-9.25)2 = 328.1875 and
Var(y) = compute product for all scenarios for sum(f(x,y)*(f(y)-E(y)2) = .1(30-6.55)2 + .25(5-6.55)2 + .50(4-6.55)2+ .15(2-6.55)2 = 61.9475.
Additionally, it branches out by incorporating funds being used on them half and half, giving us the equation:
E(ax+by) = aE(x) + bE(y) -> E(r) = E(.5x + .5y) -> E(.5x + .5y) = .5E(x) + .5E(y) = .5(9.25) + .5(6.55) = 7.9.
Which then it wants us to find the risk:
Var(ax+by) = (a^2)Var(x) + (b^2Var(y)) +2abσ<subscript>(xy), where σ<subscript(xy)> is the covariance of x and y, and the covariance equation used above.
Giving us the equation Var(.5x+.5y) = (.5^2)(328.1875) + (.5^2)(61.9475) + 2(.5)(.5)(-135.3375) = 29.865. After this I am supposed to calculate the standard deviation from the variance to give us a measure of risk, but that's not really needed, in my opinion.
So now we come to the equation I first introduced, which is where I'm stuck with finding Var(x+y) given the new sub scenario. I tried the standard deviation of the variances x and y added together, but it gave me 19.7518 which is not equal to the 119.46 of the Var(x+y) part of the equation.
I tried looking at the equation for covariance, but I just get even more confused on how to plug in the numbers as I tried and got 0. I used population variance, as i don't know if sample variance works in this scenario, and couldn't figure out how to plug in the numbers for it. σxy> = sum((xi - ux)>(yi - uy)>))/N where xi is the value in the row (e.g.; x1 = -40),where N is the number of elements (I believe 4), and u is the mean or average.
Any help would be greatly appreciated as I am stuck and confused. I will clarify as best as I can if needed. Hopefully I'm just brain farting. Thanks for your time.
. Now as in the picture, all values are given to us, but i am trying to understand where each variable is coming from. The book tells me to reference an equation in a different part of the book, which gives me the information, "we have already computed Var(s) = Var(x+y)," which references the table "Calculation of the expected value and variance for total daily sales at DiCarlo Motors", which is this table:Now here's where the problem lies. The data given to me for the problem is the following table, which is separate from the previous table. This is a bivariate probability distribution for x and y, for clarity.
In the scenario, we have found the expected values of investment return on x and y:
E(x) = compute product for all scenarios for sum(f(x,y)*f(x)) = .10(-40) + .25(5) + .5(15) + .15(30) = 9.25 and
E(y) = compute product for all scenarios for sum(f(x,y)*f(y)) = .10(30) + .25(5) + .5(4) + .15(2) = 6.55.
But, the scenario further expands to find the risk associated with investments x and y giving the equations:
Var(x) = compute product for all scenarios for sum(f(x,y)*(f(x)-E(x)2) = .1(-40-9.25)2 + .25(5-9.25)2 + .50(15-9.25)2+ .15(30-9.25)2 = 328.1875 and
Var(y) = compute product for all scenarios for sum(f(x,y)*(f(y)-E(y)2) = .1(30-6.55)2 + .25(5-6.55)2 + .50(4-6.55)2+ .15(2-6.55)2 = 61.9475.
Additionally, it branches out by incorporating funds being used on them half and half, giving us the equation:
E(ax+by) = aE(x) + bE(y) -> E(r) = E(.5x + .5y) -> E(.5x + .5y) = .5E(x) + .5E(y) = .5(9.25) + .5(6.55) = 7.9.
Which then it wants us to find the risk:
Var(ax+by) = (a^2)Var(x) + (b^2Var(y)) +2abσ<subscript>(xy), where σ<subscript(xy)> is the covariance of x and y, and the covariance equation used above.
Giving us the equation Var(.5x+.5y) = (.5^2)(328.1875) + (.5^2)(61.9475) + 2(.5)(.5)(-135.3375) = 29.865. After this I am supposed to calculate the standard deviation from the variance to give us a measure of risk, but that's not really needed, in my opinion.
So now we come to the equation I first introduced, which is where I'm stuck with finding Var(x+y) given the new sub scenario. I tried the standard deviation of the variances x and y added together, but it gave me 19.7518 which is not equal to the 119.46 of the Var(x+y) part of the equation.
I tried looking at the equation for covariance, but I just get even more confused on how to plug in the numbers as I tried and got 0. I used population variance, as i don't know if sample variance works in this scenario, and couldn't figure out how to plug in the numbers for it. σxy> = sum((xi - ux)>(yi - uy)>))/N where xi is the value in the row (e.g.; x1 = -40),where N is the number of elements (I believe 4), and u is the mean or average.
Any help would be greatly appreciated as I am stuck and confused. I will clarify as best as I can if needed. Hopefully I'm just brain farting. Thanks for your time.
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