SOA Exam P Question 318 | Variance of Discrete Multivariate Distribution

Random variables X and Y have joint distribution
X = 0 X = 1 X = 2
Y = 0 1/15 a 2/15
Y = 1 a b a
Y = 2 2/15 a 1/15
Let a be the value that minimizes the variance of X.
Calculate the variance of Y.
(A) 2/5
(B) 8/15
(C) 16/25
(D) 2/3
(E) 7/10
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[MarconyIcemanYeah]

In 4:28. "a" can't be a negative number since P[X=1, Y=0] = a. Because of that, a must be in the interval [0, 1]. I like the way you deduce that "a" is 0, however, the analysis doesn't work properly when "a" is negative. In fact, if a = -(1/5), the variance would be 0. I would suggest to sum all the probabilities in the table. You will get 4a + b + 2/5. This must be equal to 1. But it is also equal to the expected value. Therefore, you may conclude that the expected value is equal to 1 as well. Because of this, you can rewrite E[x^2] as 1 + 2a + 2/5. If you calculate the variance with that expression, the variance would be: V[x] = 1 + 2a + 2/5 - (1)^2 = 2a + 2/5. Now, since a can't be negative, it is obvious that the variance would be minimized when a equals 0.

[PolKsio]

I don't get it, that the variance increases when a increases. a appears three times in the expression of the variance once you develop it, once in 6*a, a second time in -16a² and a third time in -8ab and even in the undevelopped expression, i see it in -(4a+b+2/5)² won't that make V(X) decrease when a increases?

[luisgomez]

The explanation of why the E[x] = 1 is not clear.