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Suppose $X_1$ and $X_2$ are iid $\mathcal{N}(\mu,\sigma^2)$. Meanwhile suppose after the change of variables $X_1 = R \cos(\Theta)$, $X_2 = R \sin(\Theta)$, we have

$$X_1^2 + X_2^2 = R^2 \sim \chi_2^2 $$

the chi-squared distribution with two degrees of freedom, and $$\Theta \sim \text{Uniform}[0,2\pi]$$

where $R^2$ and $\Theta$ are not necessarily independent.

Does this imply that $\mu = 0$ and $\sigma^2 = 1$? I think the answer is yes, but one of the two ways I can think to do it is to get the marginal distribution for $\Theta$ and show it is not uniform if $\mu \neq 0$. This is not appetizing, but once $\mu = 0$ the rest is easy.

I suppose it's also "clear" from the fact that the non-central chi squared distribution is not equal to chi squared, perhaps but even that statement requires proof, or at least a probability density. Is there a slick, elementary way to do it?

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  • $\begingroup$ You don't need to use the formula for the density function of the non-central chi squared distribution to carry our your plan: you could notice that the characteristic function or mgf is wrong unless $\mu=0$ and $\sigma=1$. Or the mean and variance don't match unless, etc. But I wouldn't call these slick. $\endgroup$ – kimchi lover Oct 31 '17 at 2:57
  • $\begingroup$ I suppose matching the mean and variance is almost slick, except the variance might then be written with $X_i^4$ terms, and the mean can match if you choose $\mu$ and $\sigma$ appropriately. $\endgroup$ – Drew N Oct 31 '17 at 3:03
  • $\begingroup$ Suppose $\Theta\sim\operatorname{Uniform}(0,2\pi)$ and $R^2\sim\chi^2_2.$ Let $X_1=R\cos\Theta$ and $X_2=R\sin\Theta.$ If there is an additional assumption of indepedence of $R$ and $\Theta,$ then, as is widely known, $X_1,X_2\sim \mathrm{i.i.d.} \operatorname{N}(0,1).$ So the question appear to be this: If the marginal distributions of $R$ and $\Theta$ are as above but one somehow alters the dependence between them, can that have the effect of altering the distribution of $(X_1,X_2)$ in such a way that they remain i.i.d. and remain normally distributed, but only$\,\ldots\qquad$ $\endgroup$ – Michael Hardy Oct 31 '17 at 3:16
  • $\begingroup$ $\ldots\,$only their expectations and variances change? $\qquad$ $\endgroup$ – Michael Hardy Oct 31 '17 at 3:16
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Suppose $X_1,X_2\sim\mathrm{i.i.d.} \operatorname{N}(\mu,\sigma^2)$ and $\mu\ne0.$ Then $(X_1-\mu,X_2-\mu) = S(\cos H,\sin H)$ with $H$ uniformly distributed on $(0,2\pi).$ So we have $$ R(\cos\Theta,\sin\Theta) = (\mu,\mu) + S(\cos H,\sin H). $$ Given the uniform distribution of $H,$ ask yourself whether $\Theta$ can also be uniformly distributed. Draw the picture and the answer may become clear.

If $\mu=0$ and $\sigma\ne1,$ then $R^2\sim \sigma^2\chi^2_1.$

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  • $\begingroup$ I can see that $\Theta$ will tend to "point" more towards $(\mu, \mu)$. But that's not a proof exactly. Can it be made into one? $\endgroup$ – Drew N Oct 31 '17 at 4:02
  • $\begingroup$ @DrewN : ok, So you've got the main idea. But it may be $24$ hours before I look at this again. $\endgroup$ – Michael Hardy Oct 31 '17 at 4:53
  • $\begingroup$ Look at a unit circle centered at $(\mu,\mu)$ and draw a line from $(0,0)$ to a point on that circle, and let that point move around that circle and look at how that line changes. The slope of that line is $\tan\Theta. \qquad$ $\endgroup$ – Michael Hardy Oct 31 '17 at 5:26
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By now this is beating a dead horse, but: $R^2/\sigma^2$ is non central chi squared distributed, with (in the notation of the wikipedia article) parameters $k=2$ and $\lambda=2\mu^2/\sigma^2$, and has mean $2+\lambda$ and variance $2(2+2\lambda)$. So the expectation of $R^2$ is $(2+\lambda)\sigma^2$ and its variance is $2(2+2\lambda)\sigma^4$. So we want to know that $$(2+\lambda)\sigma^2=2\tag{1}$$ and $$2(2+2\lambda)\sigma^4=4\tag{2}$$ implies $\mu=0$ and $\sigma=1$.

Divide the square of (1) by (2) to obtain $$\frac{4+4\lambda+\lambda^2}{4+4\lambda} = 1$$ to learn $\lambda^2=0$ and hence $\mu=0$. Now substitute into (1) to learn $\sigma=1$. As mentioned in a comment, this is far from slick. But it is straightforward.

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I'm not going to accept my own answer, because I'd like to see if there's a more "computational" way of doing it. However I would like feedback on whether this is even convincing enough to obviate a separate proof, or if this the present proof is too handwavy.

We first argue that $\mu \neq 0$ is inconsistent with the assumption that $\Theta \sim \text{Uniform}[0,2\pi]$. Suppose $\mu = 0$ and consider the following two regions on the cartesian plane:

  1. $0 \leq r < \infty$ and $\pi/6 < \theta < \pi/3$

  2. $0 \leq r < \infty$ and $\pi/6 + \pi < \theta < \pi/3 + \pi$.

Suppose $f(x,y)$ is a bivariate normal pdf with zero mean and diagonal covariance matrix on these regions; then $f(x,y) = f(-x, y) = f(x, -y)$. Suppose we integrate $f$ over these two regions; they must have equal volume since they correspond to the same $x$ and $y$ values, just with signs flipped (first quadrant vs. third quadrant). Imagine these regions to be "shaded in". We will call them the old regions.

Now let us physically translate the pdf and these old "shaded in" regions along the line $y = x$, that is the angle $\theta = \pi/4$. We arrive at a new nonzero mean on the line $y = x$, and the shaded in regions are centered at this new mean. Now we repeat the integration of the above two regions, which are "new", centered at 0. However, $y = x$ bisects the translated "old" regions centered at the chosen mean as well as the "new" regions centered at 0, so one old region will be strictly contained in a new region, and one new region will be strictly contained in an old region. So the volume under the two new regions is not equal and $\Theta$ cannot be uniform on $[0, 2\pi]$.

An analogous argument holds for any desired mean vector. Just pick an angular region that is entirely inside one quadrant whose bisector crosses your desired mean. If your desired mean is on an axis then one of the entries is zero, so there is no "sign flip", and you translate along the axis.

Once $\mu = 0$ the rest is easy. We have $\mathbb{E}(X_1^2 + X_2^2) = 2 \text{Var}(X_1)$, by iid, so to match the variance of $\chi_2^2$, which is 2, we must have $\text{Var}(X_1) = 1$.

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