伽马函数极限证明请求：学生t分布收敛正态分布关键步骤

Alright, let's work through this limit proof step by step—since it's the critical piece for showing the Student's t-distribution converges to the normal distribution, we'll lean on Stirling's approximation, the standard tool for analyzing Gamma function behavior as its argument grows large.

For large $z$, the Gamma function has the asymptotic expansion:
$$\Gamma(z) \sim \sqrt{2\pi} z^{z - 1/2} e^{-z}$$
Here, $\sim$ means the ratio of the two sides approaches 1 as $z \to \infty$. This approximation is perfect for our case because we're taking $p \to \infty$, so the arguments of the Gamma functions in our limit will also grow without bound.

Let's simplify notation by setting $z = \frac{p}{2}$. As $p \to \infty$, $z \to \infty$, so we can rewrite our limit in terms of $z$:
$$\lim_{z \to \infty} \frac{\Gamma\left(z + \frac{1}{2}\right)}{\Gamma(z) \sqrt{2z\pi}}$$

First, apply Stirling's approximation to $\Gamma\left(z + \frac{1}{2}\right)$:
$$\Gamma\left(z + \frac{1}{2}\right) \sim \sqrt{2\pi} \left(z + \frac{1}{2}\right)^{\left(z + \frac{1}{2}\right) - \frac{1}{2}} e^{-\left(z + \frac{1}{2}\right)} = \sqrt{2\pi} \left(z + \frac{1}{2}\right)^z e^{-z - \frac{1}{2}}$$

Next, apply it to $\Gamma(z)$:
$$\Gamma(z) \sim \sqrt{2\pi} z^{z - \frac{1}{2}} e^{-z}$$

Now substitute these into our ratio:
$$
\frac{\Gamma\left(z + \frac{1}{2}\right)}{\Gamma(z) \sqrt{2z\pi}} \sim \frac{\sqrt{2\pi} \left(z + \frac{1}{2}\right)^z e^{-z - \frac{1}{2}}}{\sqrt{2\pi} z^{z - \frac{1}{2}} e^{-z} \cdot \sqrt{2z\pi}}
$$

Let's cancel out common terms step by step:

• The $\sqrt{2\pi}$ in the numerator and denominator cancels out.
• The exponential terms simplify to $\frac{e^{-z - 1/2}}{e^{-z}} = e^{-1/2}$.
• Combine the $z$-terms in the denominator: $z^{z - 1/2} \cdot \sqrt{2z\pi} = z^{z - 1/2} \cdot (2z\pi)^{1/2} = z^{z - 1/2} \cdot z^{1/2} \sqrt{2\pi} = z^z \sqrt{2\pi}$.

After simplifying, we're left with:
$$
\frac{e^{-1/2}}{\sqrt{2\pi}} \left(1 + \frac{1}{2z}\right)^z
$$

Now we just need to compute $\lim_{z \to \infty} \left(1 + \frac{1}{2z}\right)^z$. Remember the standard limit result:
$$\lim_{n \to \infty} \left(1 + \frac{a}{n}\right)^n = e^a$$
Here, $a = \frac{1}{2}$, so we can rewrite our term as:
$$\left(1 + \frac{1}{2z}\right)^z = \left[\left(1 + \frac{1}{2z}\right)^{2z}\right]{1/2} \to \left(e^1\right){1/2} = e^{1/2}$$

Substitute this back into our expression:
$$
\frac{e^{-1/2}}{\sqrt{2\pi}} \cdot e^{1/2} = \frac{1}{\sqrt{2\pi}}
$$

To circle back to your original goal: the Student's t-distribution's PDF includes this Gamma ratio term. As the degrees of freedom $p$ go to infinity, this ratio converges to $1/\sqrt{2\pi}$, which is the normalization constant for the standard normal distribution. This eliminates the t-distribution's heavier tails, resulting in the normal distribution as the limit.

内容的提问来源于stack exchange，提问作者Andy