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Ch 3 - Practice Question 3.6

N

nanaba

Member
Regarding the distribution of (Y1 + Y2 + Y3) - (X1 + X2 + X3 + X4). The answer sheet states N ~ (3*1200 - 4*800, 3*300^2 + 4*100^2). I expected it to be N ~ (3*1200 - 4*800, 3^2*300^2 + 4^2*100^2) due to the dimension of the variance. Please advice. Thanks!
 
nanaba said:
Regarding the distribution of (Y1 + Y2 + Y3) - (X1 + X2 + X3 + X4). The answer sheet states N ~ (3*1200 - 4*800, 3*300^2 + 4*100^2). I expected it to be N ~ (3*1200 - 4*800, 3^2*300^2 + 4^2*100^2) due to the dimension of the variance. Please advice. Thanks!
Click to expand...

Suppose \(X_1\) and \(X_2\) are independent and identically distributed as \(X\)

The short answer is that \(X_1 + X_2 \neq 2X\).

Think about it - rolling two dice and summing the score (outcomes: 2, 3, 4, ...., 12) is NOT the same as rolling one die and doubling it (outcomes 2, 4, 6, 8, 10, 12).

So the variances will not be the same.

\(var(X_1)+var(X_2) = 2var(X)\) whereas \(var(2X) = 2^2var(X)\)
 
I am fully convinced with Mr. John Lee's reply to your query
 
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