Assume $X$, $Y$ are independent zero mean random variables. Define $Z_1=X+Y$ and $Z_2=X-Y$. Then, their mean values are the same.
How does one check that $Z_1$ and $Z_2$ are not the same random variables? And how to describe their variances?
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3$\begingroup$ Your question lacks some context. Supplying it might help attract an answer. As it stands, it's not clear what your question really is. As currently posed, the answer is that $Z_1 = Z_2$ (almost surely) if and only if $Y = 0$ almost surely. $\endgroup$– cardinalCommented Jun 18, 2011 at 15:48
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The difference between $Z_1$ and $Z_2$ is $2Y$. Whenever $Y \not = 0$, they will be different. If $Y$ has a symmetric distribution about $0$ then $Z_1$ and $Z_2$ will have the same distribution as each other.
The variance of the sum of two independent random variables is the sum of their variances, as is the variance of the difference of two independent random variables, so
$$\sigma_{Z_1}^2=\sigma_{Z_2}^2=\sigma_{X}^2+\sigma_{Y}^2 $$
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2$\begingroup$ (+1) Good, concise answer. I'd be interested in hearing back from the OP that this is, indeed, the question he was seeking an answer to. $\endgroup$– cardinalCommented Jun 18, 2011 at 17:13
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$\begingroup$ Thanks. I have understand the calculation of variance.But if Y!=0, how to explain $Z_1!=Z_2$? $\endgroup$– samCommented Jun 19, 2011 at 7:52
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$\begingroup$ Can I use if Cov(X,Y)==Cov(X,X) then X=Y ? $\endgroup$– samCommented Jun 19, 2011 at 8:53
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$\begingroup$ @sam: No - in your example $Cov(X,Z_1) = Var(X)$ even though they are different. $\endgroup$– HenryCommented Jun 19, 2011 at 17:32
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$\begingroup$ Oh,so is there any way to check Z1 == Z2 or not? $\endgroup$– samCommented Jun 19, 2011 at 17:46