Give a reasonable definition of a chi-square distribution with zero degrees of freedom.
My textbook offers a hint to use the moment generating function (m.g.f.) of a distribution that is $\mathcal X^2(r)$ and let $r=0$. I have two questions regarding my approach detailed below.
Using the hint, I recall that for a random variable $X$ that is $\mathcal X^2(r)$, the m.g.f. is given by $$M_X(t) = (1 - 2t)^{-r/2}$$ Setting $r=0$, we see that $$M_X(t) = 1$$ Thus, I'm concluding that when $X$ is $\mathcal X^2(0)$, $X$ has a degenerate distribution since $$\sigma^2 = \frac{d^2lnM_X(t)}{dt^2}=0$$
Is my conclusion correct? I feel like this is an odd question to ask for an exercise.
Anyhow, if my conclusion is correct, I should be able to find the exact $x$ such that $Pr(X=x)=1$. My approach was to set $r=0$ in the following probability density function (p.d.f.)
$$ f(x) = \frac{x^{\frac r2-1}e^{-x/2}}{\Gamma(r/2)2^{r/2}} \ $$
for $0 < x < \infty$. When $r=0$, $$ f(x) = \frac{1}{xe^{x/2}} $$
Since $X$ is a degenerate distribution (based on conclusion above), there exists only one such $x$ such that $Pr(X=x) = 1$ and that $x$ is roughly equal to $0.703467$. (I'm a bit concerned that the p.d.f. magically became a p.m.f.). Is any of this valid?