Se é distribuído , é distribuído e , eu sei que é distribuído se X e Y forem independentes.
Mas o que aconteceria se X e Y não fossem independentes, ou seja,
Isso afetaria como a soma é distribuída?
Se é distribuído , é distribuído e , eu sei que é distribuído se X e Y forem independentes.
Mas o que aconteceria se X e Y não fossem independentes, ou seja,
Isso afetaria como a soma é distribuída?
Respostas:
See my comment on probabilityislogic's answer to this question. Here,
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@dilip's answer is sufficient, but I just thought I'd add some details on how you get to the result. We can use the method of characteristic functions. For anyd -dimensional multivariate normal distribution X∼Nd(μ,Σ) where μ=(μ1,…,μd)T and Σjk=cov(Xj,Xk)j,k=1,…,d , the characteristic function is given by:
For a one-dimensional normal variableY∼N1(μY,σ2Y) we get:
Now, suppose we define a new random variableZ=aTX=∑dj=1ajXj . For your case, we have d=2 and a1=a2=1 . The characteristic function for Z is the basically the same as that for X .
If we compare this characteristic function with the characteristic functionφY(t) we see that they are the same, but with μY being replaced by μZ=∑dj=1ajμj and with σ2Y being replaced by σ2Z=∑dj=1∑dk=1ajakΣjk . Hence because the characteristic function of Z is equivalent to the characteristic function of Y , the distributions must also be equal. Hence Z is normally distributed. We can simplify the expression for the variance by noting that Σjk=Σkj and we get:
This is also the general formula for the variance of a linear combination of any set of random variables, independent or not, normal or not, whereΣjj=var(Xj) and Σjk=cov(Xj,Xk) . Now if we specialise to d=2 and a1=a2=1 , the above formula becomes:
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