# Does $(X,X)'$ follow a bivariate normal distribution?

I'm fairly new to multivariate distributions. I'm trying to figure out if $$(X,X)^{'}$$ follow a bivariate normal distribution (the prime = transposed). If $$X\sim N(\mu, \sigma^{2})$$ where $$\mu \in \mathbb{R}$$ and $$\sigma^{2}\geq 0$$. What I know is that if $$\sigma^{2}=0$$ then $$X\sim N(\mu,0)$$ and then we have a distribution that has no density, hence degenerate normal distribution. But how do I conclude if it has a bivariate normal distribution in this case? And assuming that $$\sigma^{2}>0$$, then we have a normal distribution. I think my main question is how do I show that $$(X,X)^{'}$$ either has/do not has a bivariate normal distribution. Can somebody help or point me in the right direction of what I have to do?

• stats.stackexchange.com/q/240612/119261 Commented Jun 25, 2020 at 18:46
• @StubbornAtom Thank you. I see my post and the one you referred to are quite similar. I'll still leave this post because it might be helpful, with the explanation jld provided, to others. It gives an intuitive explanation of the problem for a more general case. Commented Jun 25, 2020 at 20:38

Let $$A = {1\choose 1}$$ and note that $$A$$ is a linear transformation from $$\mathbb R$$ to $$\mathbb R^2$$. Then if $$X\sim\mathcal N(\mu,\sigma^2)$$ we'll have $${X\choose X} = AX \sim \mathcal N(A\mu, \sigma^2 AA^T)$$ since linear transformations of Gaussians are also Gaussian. The covariance matrix is rank $$1$$ when $$\sigma>0$$ and rank $$0$$ when $$\sigma = 0$$ so either way this distribution will not have a Lebesgue pdf. It'll be entirely supported on $$\{x \in \mathbb R^2 : x_1 = x_2\}$$ when $$\sigma>0$$ and if $$\sigma = 0$$ we'll have $$AX$$ as a point mass at $$A\mu = {\mu\choose \mu}$$. But it is a valid bivariate Gaussian.

If you want to be more sure, we can use characteristic functions (CFs). Let $$\varphi_Y(t)$$ be the CF of a random variable $$Y$$ evaluated at $$t$$. Then $$\varphi_{AX}(t) = \text E(\exp(it^TAX)) = \varphi_{X}(t^TA) \\ = \exp\left(i\mu t^TA - \frac 12\sigma^2 (t^TA)^2 \right) \\ = \exp\left(it^T(A\mu) - \frac 12\sigma^2 t^TAA^Tt \right)$$ which is the CF of a $$\mathcal N(A\mu, \sigma^2 AA^T)$$ random variable.

This is also a reason for defining a multivariate Gaussian as a random variable $$X\in\mathbb R^n$$ where $$a^TX\sim\mathcal N(a^T\mu,a^T\Sigma a)$$ for all $$a\in\mathbb R^n$$ rather than by specifying the pdf, since this case includes Gaussians with singular covariance matrices for which there is no Lebesgue pdf.

• Thank you so much for your answer. The problem is much clearer to me now. Just to be sure we can conclude that since there is no Lebesgue pdf then $(X,X)'$ is not absolutely continuous wrt. the Lebesgue measure on $\mathbb{R}$? Commented Jun 25, 2020 at 20:32
• @Leonardo yeah exactly, the set $H := \{x\in\mathbb R^2 : x_1=x_2\}$ has measure 0 w.r.t. the Lebesgue measure on $\mathbb R^2$ yet $P((X,X)^T \in H) = 1$ so there's no Radon-Nikodym derivative of these measures here
– jld
Commented Jun 25, 2020 at 20:34
• Perfect, it is exactly what I've been trying to figure out. Once again, thank you for your explanation Commented Jun 25, 2020 at 20:40
• @Leonardo you’re welcome, glad this helped!
– jld
Commented Jun 25, 2020 at 20:41