I have the following expression:

$$ Z = G - N + E $$


$$ G \sim \mathrm{Gamma}(k,\frac{\sigma^2}{k}) \\ N \sim \mathcal{N}(0,\dfrac{\alpha{}\sigma^2}{k^{2}}) \\ E \in \mathbb{R}_{+} $$

This expression is obtained when the estimator of the variance for an IID normal distribution was used under the presence of a signal. More specifically,

$$ Z = \dfrac{\sigma^2}{2k}\sum_{n=0}^{k-1}U_{n}^{2} - \dfrac{\sqrt{\alpha\sigma^2}}{k}\sum_{n=0}^{k-1}U_{n}s[n] +\dfrac{\alpha}{2k} $$

with $U \sim \mathcal{N}(0,1)$ and $\mathbf{s}^{T}\mathbf{s} = 1$.

I know that the expected value is a linear operator (i.e. $\mathbb{E}[X + Y] = \mathbb{E}[X] + \mathbb{E}[Y]$). The variance isn't a linear operator but it allows to decompose the expression further, to an extent.

Does this apply, however, in the case when the distributions are not similar? I have seen from some Googling that under the assumption of independence, the joint PDF was derived instead. I'm not sure if the two distributions are independent.

  • $\begingroup$ you will need to calculate terms $E(XY)$. Íf the independence is not known you will not be able to solve this. If independent use $E(XY) = E(X)E(Y)$. $\endgroup$ – Stefan Jun 12 '18 at 15:31
  • $\begingroup$ Out of curiosity, how can I determine if the two r.v. are independent? $\endgroup$ – KaiserHaz Jun 12 '18 at 15:57
  • $\begingroup$ you need som kind of model describing the interaction, or else you need to show that the probability distribution factorizes e.g. $P(X\in A,Y\in B)=P(A)P(B)$ if you have a probabiity denity function then that should factorise, if one has a density and the other is discrete it is enough that $P(X=i,Y\in [x,x+dx]) = p_X(i)f_Y(x)dx$ I think. $\endgroup$ – Stefan Jun 12 '18 at 16:22
  • $\begingroup$ @Stefan I edited the question with a bit more details on the variables. $\endgroup$ – KaiserHaz Jun 13 '18 at 5:48
  • $\begingroup$ there are missing indicies $\endgroup$ – Stefan Jun 13 '18 at 8:04

What needs analysis to be calcuated is $E(G N)$, but they can be deduced from the model to lead to the need to evaluate terms of the kind $c_{ij}E(U_i^2U_j)$ which is zero because all odd moments of the standard normal distribution is zero. hence

$$ Var(G+N) = E((G+N)^2) - ((E(G+N))^2 = E(G^2) + E(N^2) - ((E(G+N))^2 $$

  • $\begingroup$ Can the final term $E(G+N)$ be distributed to $E(G) + E(N)$? $\endgroup$ – KaiserHaz Jun 13 '18 at 8:17
  • $\begingroup$ yes that is valid for all $G,N$ becase integration is linear. $\endgroup$ – Stefan Jun 13 '18 at 8:20

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