I've been given a problem where I have

$$ X \sim \mathcal{N}(\mu = 2, \sigma^2 = 9) $$

$$ Y \sim \mathcal{N}(\mu = 3, \sigma^2 = 4) $$

Their correlation is $ \rho_{XY} = 0.6 $. First I am asked for the mean and standard distribution of X+Y, which I found as

$$ \mu_{X+Y} = \mu_X + \mu_Y = 2 + 3 = 5 $$

$$ \sigma_{X+Y} = \sqrt{\sigma_{X}^2 + \sigma_{Y}^2 + 2 \rho_{XY} \sigma_x \sigma_y} $$

However, I am also asked what the distribution of $ X + Y $ is, what if $X$ and $Y$ are jointly normally distributed, and what if $X$ and $Y$ are not jointly normally distributed. I am unsure how I would go about determining what the distribution of $X + Y$ is, though I assume that

  • If $X$ and $Y$ are jointly normally distributed then $X + Y$ is normally distributed

  • If $X$ and $Y$ are not jointly normally distributed then $X + Y$ is not normally distributed

How do I determine what the distribution of $X + Y$ is when I am not told whether they are jointly distributed?


I read the following post, but wasn't sure how to apply what was said to my current problem, if I could at all.

If $X$ and $Y$ are normally distributed random variables, what kind of distribution their sum follows?

  • $\begingroup$ An important equation on the linked page is $\sigma_{X+Y}^2 = \sigma_{X}^2 + \sigma_{Y}^2 + 2\operatorname{cov}(X,Y).$ This question has more the flavor of a 'textbook exercise' rather than of a practical application. $\endgroup$
    – BruceET
    Commented Sep 23, 2018 at 20:41
  • $\begingroup$ Don't I already have that in the standard deviation calculation in my post though @BruceET ? $\endgroup$
    – strwars
    Commented Sep 23, 2018 at 20:43
  • 2
    $\begingroup$ Indeed. But you said you don't see the relevance of the link. Perhaps a key statement in common is a clue to that. // Marginals of a bivariate normal distribution are normal. But a distribution that is not bivariate normal can have normal marginals. $\endgroup$
    – BruceET
    Commented Sep 23, 2018 at 20:45

1 Answer 1


Comment (continued), illustrating the last sentence of my Comment above with a simulation.

For $X$ and $Y,$ let means be 0, SDs 1, and $\rho = 0.8.$ A sample of size 100,000 from their joint binomial distribution is simulated below in R. [See this page for the method of simulation.]

m = 10^5;  rho = .8; phi = sqrt(1-rho^2)
x = rnorm(m);  z = rnorm(m)
y = rho*x + phi*z
[1] 0.7997351  $ aprx 0.8

The bivariate normal distribution is suggested by the upper-right plot in the figure below. The two histograms in the top row suggest the standard normal marginal distributions.

Now, manipulate the results so that the marginals are unchanged, but the joint distribution is no longer bivariate normal, and the sum $T = X + Y_1$ is not normal.

y1 = y;  
 y1[x < 0] = -abs(y[x < 0])
 y1[x >= 0] = abs(y[x >- 0])
[1] 0.8546086  # > 0.8

enter image description here

Finally, what would happen if $Y_2 = -X?$ Then are $X$ and $Y_2$ bivariate normal? Is $X + Y_2$ normal?

  • $\begingroup$ I'm a little bit lost. So the only way to determine X+Y's distribution in my problem is to generate a bunch of values using the distribution as the generator and then plotting the results in a histogram? Also, even if both jointly normal distributions and non-jointly normal distributions can have normal marginals, I'm not sure how that relates to what I said in my post. I'm not trying to be rude and am very appreciative of your help, I'm just not really understanding it. $\endgroup$
    – strwars
    Commented Sep 23, 2018 at 23:49
  • $\begingroup$ Not suggesting simulation as an alternative to understanding the theory. Just showing possibilities to try to get you to think clearly and productively. The two bullets in your Question are more complicated than may be immediately obvious. For the first part, maybe tomorrow you should give the link another chance. It is very well written. $\endgroup$
    – BruceET
    Commented Sep 24, 2018 at 0:10
  • 1
    $\begingroup$ I re-read the page and appreciate what it was saying. Changing my perspective on my second bullet. Thanks! $\endgroup$
    – strwars
    Commented Sep 28, 2018 at 5:12

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