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1

Assume $Y = g(X)$ be the function of RV $X$, then by using the following $$E\{ g(X) \} = \int g(x) f_X(x) dx$$ variance of $Y$ can be computed without the computation of pdf $f_Y(y)$ as: \begin{align} \text{Var(Y)} &= E\{ (Y-\mu_Y)^2 \} = E\{ Y^2 \} - (\mu_Y)^2 \\ & = E\{ g^2(X) \} - E\{ g(X) \}^2 \\ & = \int g^2(x) f_x(x) dx - \left(\... 0 The relationship \mathbb{E}[XY]=\mathbb{E}[X]\mathbb{E}[X] says that the two random variables are uncorrelated. Uncorrelated random variables are in general not independent, unless they are Gaussian random variables. 1 Let's look at things in the following perspective; Statistical independence is an assumption imposed on events A,B or random variables X,Y, whose (the assumption's) mathematical consequence is expressed as; P( A \cap B) = P(A) P(B)$$and$$ E\{X Y\} = E\{X\}E\{Y\}  So these are the consequences of independence, but not the cause of it. (This is ...

1

Essentially $E\{XY\}=0$ (assuming $E\{X\}=E\{Y\}=0$) means that $X$ and $Y$ are uncorrelated, but uncorrelated is a weaker property than probabilistic independence. A collection of independent random variables will always be uncorrelated, but being uncorrelated does not mean two random variables are independent.

3

No, NO and $\mathbf{NO}$. Knowing that $E[X]=E[Y]=0$ and $E[XY]=0$ (or more generally that $E[XY]=E[X]E[Y]$ or equivalently, $E[XY]-E[X]E[Y]=0$) does not help in the least in proving or deducing that $X$ and $Y$ are independent random variables. The conditions stated in the previous sentence are necessary for $X$ and $Y$ to be deemed independent random ...

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