Let $x_n, y_n$ be sequences of zero mean random variables, not necessarily i.i.d. Suppose that there are finite $\sigma_1^2,\sigma_2^2$ such that $$x_n\overset{d}{\to} N(0,\sigma_1^2), $$ and $$y_n\overset{d}{\to} N(0,\sigma_2^2). $$ Can I say that $x_n+y_n{\to} N(0,\sigma^2)$ for some finite $\sigma^2$?

I know that I can't describe $\sigma_2$, but the limiting distribution of $x_n+y_n$ is still normally distributed?

  • $\begingroup$ Are $x_n$ and $y_n$ independent? $\endgroup$ May 20, 2020 at 22:12
  • $\begingroup$ no. it is not needed $\endgroup$ May 20, 2020 at 23:05
  • $\begingroup$ Please add that as an edit to the question. Not everybody reads comments. $\endgroup$ May 21, 2020 at 1:15
  • $\begingroup$ You can say it if you like but whether it is a true statement or not is a different matter. The sum of two normal random variables is not a normal random variable unless the variables are jointly normal (which joint normality would hold for independent random variables but you have ruled it out). $\endgroup$ May 21, 2020 at 1:16
  • $\begingroup$ @DilipSarwate Yes. I perceived that the limiting distribution of $x_n+y_n$ need not to be Gaussian. $\endgroup$ May 21, 2020 at 2:09

1 Answer 1


Even in the special case where $X_n \sim N(0,\sigma_{1,n}^2), Y_n \sim N(0,\sigma_{2,n}^2)$ where $\{\sigma_{1,n}^2\}$ and $\{\sigma_{2,n}^2\}$ are sequences of positive real numbers converging to $\sigma_1^2$ and $\sigma_2^2$ respectively, and so $X_n \overset{d}{\to}N(0,\sigma_{1}^2), Y_n \overset{d}{\to} N(0,\sigma_{2}^2)$, it is not possible to assert that $X_n+Y_n$ is a normal random variable of any kind or that $X_n+Y_n$ converges in distribution to a normal random variable unless it is also asserted that $X_n, Y_n$ are jointly normal (which implies that $X_n, Y_n$ are also individually (marginally) normal random variables). If $X_n$ and $Y_n$ are indeed jointly normal with correlation coefficient $\rho_n$ where $\lim_{n\to\infty} \rho_n = \rho$, then $$X_n+Y_n \overset{d}{\to}N(0,\sigma_{1}^2 + \sigma_{2}^2 + 2\rho \sigma_{1}\sigma_{2})$$


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