Proving Mean and Variance Equivalence: Step-by-Step Guide | Helpful Tips

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To prove the equivalence of mean and variance, it's essential to specify the distribution in question. The discussion highlights that if Y is defined as aX, the mean can be calculated using E[Y] = E[aX] = aE[X], while the variance is given by Var[Y] = Var[aX] = a^2Var[X]. Clarifying the context and the specific distribution is crucial for accurate calculations. Participants are encouraged to provide more details for tailored assistance. Understanding these relationships is key to proving the mean and variance equivalence.
mcguiry03
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μ=αθ
σ^2=αθ^2
can someone show me how to prove that the mean and variance is equal to what is at the top... tnx a lot...:biggrin:
 
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Hey mcguiry03 and welcome to the forums.

You need to specify what the distribution is and what you are trying to do.

If you are starting with a distribution X and you have Y =aX and want to find the mean and variance of Y then you use the fact that E[Y] = E[aX] = aE[X] and Var[Y] = Var[aX] = a^2Var[X].

If it is something else, then you will need to specify what is going on and what you are trying to do.
 
tnx sir
 

Thread 'Formal derivation of statement from Peano Arithmetic system'

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...
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