Variance Properties: Understanding Var(aX+bY+c) & Solving for Var(aX+bY)

AI Thread Summary
The variance of a linear combination Var(aX + bY + c) is indeed equal to Var(aX + bY), as adding a constant does not affect variance. The formula Var(aX + bY) can be applied, which is a^2Var(X) + b^2Var(Y) + 2abCov(X,Y). This confirms the initial reasoning presented in the discussion. The conclusion reached is accurate, and the topic is resolved.
ja404
Messages
10
Reaction score
0
[SOLVED] Properties of variance

Would the Var(aX + bY + c) just be the Var(aX+bY) since adding a single number to the function doesn't change the variance. I would then be able to use the property:

Var(aX+bY)= a^2Var(X)+b^2Var(Y)+2abCov(X,Y)

Just wondering if anyone can confirm my reasoning here. Thanks.
 
Physics news on Phys.org
Correct.
 
Thank you, that's all I need. Mods can close.
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
Back
Top