Negative log of unit scale gamma distribution converges to standard exponential

Question

Let $Y \sim \text{Gamma}(\alpha,1)$, with pdf $f(y) = y^{\alpha-1}e^{-y}/\Gamma(\alpha)$, where $\alpha \in (0,1)$. Let $Z = -\alpha \log Y$. Prove that $Z$ converges in distribution to a standard exponential random variable as $\alpha \to 0$.

The characteristic function of $Z$ is

$\phi_Z(t) = E e^{itZ} = E Y^{-i \alpha t} = \frac{\Gamma(\alpha - i\alpha t)}{\Gamma(\alpha)}$

The last term should converge to $1/(1-it)$, but why is this true?

Answer 1

Answering this myself for posterity. The final term here is

$$\frac{\Gamma(\alpha - i\alpha t)}{\Gamma(\alpha)} = \frac{\alpha(1-it)\Gamma(\alpha(1 - i t))}{\alpha(1-it)\Gamma(\alpha)} = \frac{\Gamma(\alpha(1-it)+1)}{(1-it)\Gamma(\alpha+1)} \to \frac{1}{1-it}$$ as $\alpha \downarrow 0$.

Answer 2

Why not simply compute the distribution of $Z$ explicitly via transformation? Let $Z = g(Y) = -\alpha \log Y$ or $Y = g^{-1}(Z) = e^{-Z/\alpha}$. Then

$$f_Z(z) = f_Y(g^{-1}(z)) \left|\frac{dg^{-1}}{dz}\right| = \frac{1}{\Gamma(\alpha)} \left(e^{-z/\alpha}\right)^{\alpha-1} \exp(-e^{-z/\alpha}) \cdot \frac{1}{\alpha} e^{-z/\alpha} = \frac{\exp(-e^{-z/\alpha} - z)}{\Gamma(\alpha+1)}.$$ Then as $\alpha \to 0^+$ we clearly have $$f_Z(z) \to e^{-z}.$$