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$X\sim N(52,6)$, $Y\sim (40,8)$. What's the standard deviation of $Z=X+Y$?

I'm considering to transform the linear relationship to matrix form $$Z=\begin{pmatrix} 1& 1\\ \end{pmatrix}\begin{pmatrix} X\\ Y \end{pmatrix}$$

and apply $$\Sigma_{z}=A \Sigma^{-1} A'$$

However, I don't know the next step. How to calculate $\Sigma$?

Thank you!

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    $\begingroup$ A hint: In case of independent random variables it is known that characteristic function of the sum $Z=X+Y$ is equal to the product of characteristic functions of separate variables: $$\phi_{Z=X+Y} (t) = \phi_{X}(t)*\phi_{Y}(t)$$. Can you infere the family of distributions from resulting characteristic function? $\endgroup$ – Tomas Oct 24 '12 at 9:48
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    $\begingroup$ @Thomas I don't think that this simple case requires convolution $\endgroup$ – jem77bfp Oct 24 '12 at 10:01
  • $\begingroup$ Hint (and justification of the remark by @jem77bfp): if $X$ and $Y$ are independent random variables with any distributions for which $X$ has a standard deviation of $6$ and $Y$ has a standard deviation of $8$, then you can deduce the SD of $X+Y$ from this information alone. $\endgroup$ – whuber Oct 24 '12 at 14:05
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As distribution of $\begin{pmatrix} X\\ Y \end{pmatrix}$ is not specified, I would probably assume that $\mbox{cov}(X,Y)=0$ then $$\Sigma =\begin{pmatrix} 6 & 0\\ 0 & 8 \end{pmatrix}.$$ Where diagonal elements of $\Sigma$ are variances of $X$ and $Y$. If the $\mbox{cov}(X,Y)$ is available you should insert that number instead of $0$'s.

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It's interesting that you are trying to use the matrix form when this is more of a elementary problem.

Remember $$ \mbox{Var}(X + Y) = \mbox{Var}(X) + \mbox{Var}(Y) + 2 \mbox{Cov}(X,Y). $$

You don't say anything about the relation between $X$ and $Y$, so either you assume a covariance of zero or the problem is not solvable due to missing informations.

The matrix form of this is $$ \mbox{Cov}(AX) = A \cdot \mbox{Cov}(X)\cdot A'. $$

This covariance matrix seems to be your $\Sigma$. I don't know why you inverted it in your formula. Perhaps I misunderstand your equation? You know that $\Sigma_{1,1} = 6$ and $\Sigma_{2,2} = 8$. These are just the variances of $X$ and $Y$ respectively. Again, since you do not have a covariance and since you cannot calculate it from the information given, the question is either unsolvable or you have have to assume independence, which implies $\Sigma_{1,2}=\Sigma_{2,1}=0$. This will then give the same result as the equation for $\mbox{Var}(X+Y)$. Remember to take the root to get the standard deviations.

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  • $\begingroup$ Why do you have a minus sign in front of the 2? $\endgroup$ – mark999 Oct 24 '12 at 10:07
  • $\begingroup$ @Erik "$\cdot$" denotes multiplication and "$*$" denotes convolution. $\endgroup$ – jem77bfp Oct 24 '12 at 10:18
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If $X$ and $Y$ are independent random variables that are normally distributed, then their sum is also normally distributed. i.e., if

$X\sim N(\mu_X, \sigma_X^2)$

$Y \sim N(\mu_Y, \sigma_Y^2)$

$Z=X+Y$

and $X$ and $Y$ are independent, then

$Z \sim N(\mu_X + \mu_Y, \sigma_X^2 + \sigma_Y^2).$

You can prove it using both characteric functions and convolutions. I will refer you to read this page for the proofs.

In the event that the variables $X$ and $Y$ are jointly normally distributed correlated random variables, then $Z=X + Y$ is still normally distributed and the mean is the sum of the means. However, the variances are not additive due to the correlation. Indeed,

$\sigma_{X+Y} = \sqrt{\sigma_X^2+\sigma_Y^2+2\rho\sigma_X \sigma_Y}$

This one is also shown in the previous link.

You can also find this using the fact that $Q^{'}=[X \; Y]$ is bivariate normal with variables $X$ and $Y$ and distributed as $N_{2}(\mu,\Sigma).$ Then $Z=AQ$ with $A=[1\;1]$ is distributed as $N_{q}(A\mu,A \Sigma A^{'})$, where $q$ is the number of linear combinations (or the number of rows in $A$). Each row in $A$ corresponds to a linear combination of the variables $X$ and $Y$. Here we have only 1 row. So, $q=1$. Put $\sigma_X^2$, $\sigma_Y^2$ and $\sigma_{XY}$ in $\Sigma_{(2\times2)}$ and find the result for the correlated case.

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    $\begingroup$ Your first statement is not quite correct. If $X$ and $Y$ are independent normal random variables, then they necessarily are jointly normal random variables and the formula used in the second part of your answer applies to them too, and gives the same results as you stated separately because $\rho = 0$. A proof of the result for jointly normal random variables can be found here on-site without needing to go to Wikipedia. $\endgroup$ – Dilip Sarwate Oct 24 '12 at 14:29
  • $\begingroup$ Probably the choice of words was bad! I wanted to mean that their joint density need not be considered to find the distribution of $Z$. Thank you for mentioning. $\endgroup$ – Blain Waan Oct 24 '12 at 16:33
  • $\begingroup$ Maybe you could take the time to edit your answer. Note that "...their joint density need not be considered to find the distribution of $Z$." is also not a correct statement. What is correct are the comments by jem77bfp and whuber on the main question and the answer by Erik that normality or joint normality or the lack thereof are irrelevant to the computation of the variance (or standard deviation) of $Z = X+Y$ as long as independence is assumed or as long as $\text{cov}(X,Y)$ is known. $\endgroup$ – Dilip Sarwate Oct 24 '12 at 19:42

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