Okay I know to get V(Y) for Y = X1 + X2 you just need to do V(Y) = a^2(X1) + b^2(X2). To get the standard deviation you need to take the square root of V(Y).
The only problem is, I don't know what to do when you have coefficients in front of the X1 and X2, and especially coefficients like 4 and 6, so it looks like Y = 4X1 - 6X2 so that solving for V(Y) makes it come out negative (since you're subtracting a larger number from a lower number). And apparently Variance can't be negative...
What should I do in this situation? Am I solving it incorrectly? X1 and X2 are independent Bernoulli distributions btw... if someone can please explain how to do something like V(Y) = 16X1 - 24X2 to me using any proportion or value for p that would be greatly appreciated. Or just tell me if I should make the variance from negative to positive and then take the square root of that to find the standard deviation. Thank you so much!
The only problem is, I don't know what to do when you have coefficients in front of the X1 and X2, and especially coefficients like 4 and 6, so it looks like Y = 4X1 - 6X2 so that solving for V(Y) makes it come out negative (since you're subtracting a larger number from a lower number). And apparently Variance can't be negative...
What should I do in this situation? Am I solving it incorrectly? X1 and X2 are independent Bernoulli distributions btw... if someone can please explain how to do something like V(Y) = 16X1 - 24X2 to me using any proportion or value for p that would be greatly appreciated. Or just tell me if I should make the variance from negative to positive and then take the square root of that to find the standard deviation. Thank you so much!
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Variance is never negative. In your example, Y=4X1-6X2, V(Y)= 16V(X1) + 36 V(X2).
always add the variances of the terms
always add the variances of the terms