I'm stuck with this question: Suppose I have two random variables: A and B such that $$A\sim N(\mu_A,\sigma_A)$$ $$B\sim N(\mu_B,\sigma_B)$$

A and B are independent.

I create a new random variable $Y=A+B$ How would the joint pdf of A and Y be? i.e. is there an explicit form for $$f_{A,Y}(a,y)$$ where the small letters denote the realised values.

My issue is since Y contains A, I can't just write the pdf as product of two pdf right?


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    $\begingroup$ The dependence, if any, between A and B is crucially important. If you don't understand why, you need to learn the basics of 1st semester undergrad probability theory. Do you even understand what joint distribution and joint pdf are? $\endgroup$ – Mark L. Stone Jun 14 '15 at 21:06
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    $\begingroup$ Do you understand what independence of random variables is, and how that relates to their joint distribution and, if it exists, to their joint pdf? $\endgroup$ – Mark L. Stone Jun 14 '15 at 21:14
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    $\begingroup$ See here for why simply specifying the marginals distributions is not sufficient. There are numerous posts on this topic -- see also here for example, which shows another example indicating why identifying both the marginal distributions and the correlation is still not sufficient. $\endgroup$ – Glen_b Jun 15 '15 at 1:56
  • $\begingroup$ Anyone who followed my advice in stats.stackexchange.com/questions/33776/… , and read at least the beginning of Feller Vol. 2, would of course be familiar with the examples in problems 2 and 3 of III.9 and problems 5 and 7 of V.12, of random variables whose marginal distributions are Normal but which are not jointly normal. $\endgroup$ – Mark L. Stone Jun 15 '15 at 4:30
  • $\begingroup$ And en.wikipedia.org/wiki/… for an example of uncorrelated normals which are nonlinearly dependent and not jointly normal. I.e.,, uncorrelated does not mean independent, even for Normals, unless they are jointly Normal. $\endgroup$ – Mark L. Stone Jun 15 '15 at 4:38

According to Hamedani, G. G. (1992). Bivariate and multivariate normal characterizations: a brief survey. Communications in Statistics-Theory and Methods, 21(9), 2665-2688), the oldest characterization of the bivariate normal distribution is due to Cramer (1941). It states (using the OP's notation)

The random vector $[Y, A]$ has a bivariate normal distribution if and only if every linear combination of $Y$ and $A$ has a univariate normal distribution.


$$\mathbf z = (z_1, z_2)' \in \mathbb R^2$$

and consider

$$W(\mathbf z) = [Y, A]\cdot \mathbf z = Yz_1 + Az_2 = (A+B)z_1 + Az_2 = (z_1+z_2)A +z_1B$$

By assumption, $A$ and $B$ are normal random variables. Scaling does not affect normality. Therefore, $(z_1+z_2)A$ is a normal random variable, and so is $z_1B$. Moreover, $A$ and $B$ are assumed independent, and therefore so are these scaled versions of them. The sum of two independent normal random variables is a normal random variable. So $W(\mathbf z)$ is a normal random variable, for any $\mathbf z$ (even for $\mathbf z = \mathbf 0$, see comments). But this means that every linear combination of $Y$ and $A$ has a univariate normal distribution, so Cramer's condition is satisfied and $[Y,A]$ has a bivariate normal distribution.

So the joint density will be the bivariate normal density, and the only thing one needs to calculate is the correlation coefficient between $Y$ and $A$, which is trivial.

  • $\begingroup$ The restriction $\mathbf{z \neq 0}$ is not included in Cramer's definition and so should not be included here either. If $\mathbb{z=0}$, $W(\mathbf z) = 0$ is a normal random variable with zero variance. For a vigorous discussion as to whether a constant can be considered to be a normal random variable with zero variance, see the comments following this answer to a different question. $\endgroup$ – Dilip Sarwate Jun 16 '15 at 12:01
  • $\begingroup$ @DilipSarwate You are right of course, and I adjusted the answer. In my opinion "constants as random variables" is one of those not-so-importnat-things that can (should?) be left out when an important result is to be communicated to a broader, and perhaps not so experienced, audience. $\endgroup$ – Alecos Papadopoulos Jun 16 '15 at 13:57

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