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I am aware that the sum of two (or more) normally distributed Random Variables is not necessarily also normal.

I do have questions regarding the special case where I have the following addition:

For c $\in \mathbb R$, is c + N(0,a), where I add a normal Random Variable with mean zero and variance a to c, is then the sum normally distributed? Also, does this addition correspond to drawing a random variable X too from a Normal distribution with mean and then add it to N(0,a)?

Hope the question is clear, thanks

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I am aware that the sum of two (or more) normally distributed Random Variables is not necessarily also normal.

Yes, it's usually only the case if they're jointly normal (multivariate normal)

For c ∈R, is c + N(0,a), where I add a normal Random Variable with mean zero and variance a to c, is then the sum normally distributed?

Yes.

Also, does this addition correspond to drawing a random variable X too from a Normal distribution with mean and then add it to N(0,a)?

You mean, if $Y\sim N(0,a)$ and $X\sim N(\mu_X,\sigma^2_x)$? The previous result means that $X+Y|X=c$ is normal. That's useless when you don't condition on the value of $X$, though, and then the form of the dependence between $X$ and $Y$ that you started with again comes in.

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  • $\begingroup$ I see. So then c + N(0,a) is normally distributed with N(c,a) - is this correct? What is the conditioning you mention in this case? Not sure I understand that part $\endgroup$
    – TestGuest
    Apr 4, 2013 at 10:04
  • $\begingroup$ "So then c + N(0,a) is normally distributed with N(c,a)" Yes. The conditioning is conditioning on $X$ taking the value $c$. en.wikipedia.org/wiki/Conditional_probability_distribution; these kinds of ideas are covered in pretty much any basic college level mathematical statistics/probability text. $\endgroup$
    – Glen_b
    Apr 4, 2013 at 10:58
  • $\begingroup$ I know about conditioned probabilities, but I had never read that sums of random variables can be understood as conditioned probabilities too...did not think that was covered in any basic text. $\endgroup$
    – TestGuest
    Apr 4, 2013 at 21:32
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    $\begingroup$ Your conclusion that 'sums of random variables can be understood as conditioned probabilities' is ... pretty much a misunderstanding of what was going on. A conclusion based on adding a constant to a random variable carries over to adding two random variables if one of them is held constant. That 'holding one constant' is what introduces conditional distributions. If you don't hold it constant, the conclusion based on adding a constant simply doesn't apply. It's not complicated or mysterious - it's a result of adding constants and attempting to generalize to adding random variables. $\endgroup$
    – Glen_b
    Apr 4, 2013 at 21:49

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