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

  • Thread starter mcguiry03
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In summary, the conversation discusses how to prove that the mean and variance of a distribution are equal to the values given at the top, μ = αθ and σ^2 = αθ^2. The process involves finding the mean and variance of a new distribution Y, which is equal to aX, and using the properties of expected value and variance. More information is needed to determine the specific distribution and problem being solved.
  • #1
mcguiry03
8
0
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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  • #2
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.
 
  • #3
tnx sir
 

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