I have a distribution described by the density function 2x*exp(-x^2). I would like to get the distribution that would by produced by taking the average of n observations drawn from this original distribution. I think eventually this will converge to a normal distribution via the central limit theorem, but want to get an idea of the shape of the distribution in the intermediate cases.
5
-
1$\begingroup$ 1. How does this question arise?/What's it for? Do you need an algebraic solution or would a numeric solution (enough to draw an accurate picture of the density) be sufficient? $\,$ 2. The density needs a some statement of the values of $x$ over which your random variable has support, presumably $x>0$. $\,$ 3. This is then the density of a Rayleigh distribution with scale $σ=√\frac12$. It's also a scaled version of a chi distribution (NB not chi-squared) with 2 d.f.; these names may help if you're trying to find information. $\endgroup$– Glen_bCommented Jan 5 at 3:35
-
1$\begingroup$ The case for the sum of $n=2$ i.i.d. Rayleigh variables (from which the mean is straightforward rescaling) is discussed here: stats.stackexchange.com/questions/414195/… . Note that even at $n=2$ the density is not available in closed form but it can be evaluated using special functions. More generally, there's ways to do numerical convolution that can give the density over a grid of values. $\endgroup$– Glen_bCommented Jan 5 at 3:40
-
1$\begingroup$ Your title is much more general than the body of your post, and has a different answer from the question in the body text. Please consider using a title more closely reflecting the much more specific question in the body of your post. $\endgroup$– Glen_bCommented Jan 5 at 3:53
-
$\begingroup$ This is not a density function, because it's negative when $x\lt 0.$ Surely you intend that $x\ge 0$! When that's assumed, the characteristic function of the density is $$\phi(t)=1 - \sqrt{\pi} ti/2 - t^2/2 + \sqrt{\pi} t^3i/8 + t^4/12 +\cdots$$ and, to assess its shape, the rest is easy. $\endgroup$– whuber ♦Commented Jan 5 at 15:29
-
1$\begingroup$ Thanks @Glen. I stuck it back in there. $\endgroup$– whuber ♦Commented Jan 6 at 18:15
Add a comment
|