借助伽马函数推导正态分布方差的积分求解疑问

Hey there! Let's work through this together—your question points to a common pitfall when using the gamma function to derive the normal distribution's variance, so let's break it down step by step.

First, let's spot a critical red flag: the integral you've ended up with—$$2\int_{-\infty}^\infty ue^{-u}du$$—actually diverges (it doesn't resolve to a finite value). That's a clear sign there's a misstep in your variable substitution, because the variance of a normal distribution is always a finite, positive number. Let's correct that and tie it back to the gamma function properly.

Why Your Current Integral Fails #

Let's split your integral into two parts to see why it doesn't converge:

• Split into: $$\int_{-\infty}^0 ue^{-u}du + \int_0^\infty ue^{-u}du$$
• For the left-hand integral, substitute $v = -u$ (so $u = -v$, $du = -dv$). When $u \to -\infty$, $v \to \infty$; when $u=0$, $v=0$. This transforms the integral to:
  $$\int_{\infty}^0 (-v)e^{v}(-dv) = \int_{\infty}^0 ve^v dv = -\int_0^\infty ve^v dv$$
• The integral $\int_0^\infty ve^v dv$ blows up to infinity, so the entire original integral can't produce a finite result. This means your substitution to reach this point was incorrect.

Correct Substitution for Normal Variance (Using Gamma Function) #

Let's start fresh with the standard normal distribution ($\mu=0$, $\sigma=1$), since scaling to non-standard normals is straightforward later. The variance is:
$$\text{Var}(X) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty x^2 e^{-\frac{x2}{2}}dx$$

Here's how to map this to the gamma function ($\Gamma(s) = \int_0^\infty u^{s-1}e{-u}du$):

1. Use evenness: The integrand $x^2 e^{-\frac{x2}{2}}$ is even, so we can rewrite the integral as:
   $$\frac{2}{\sqrt{2\pi}} \int_0^\infty x^2 e^{-\frac{x2}{2}}dx$$
2. Substitute to match gamma form: Let $u = \frac{x^2}{2}$. This gives $x = \sqrt{2u}$, and $dx = \frac{du}{\sqrt{2u}}$.
3. Plug into the integral: Substitute these values in, and simplify:
   $$\frac{2}{\sqrt{2\pi}} \int_0^\infty (2u) e^{-u} \cdot \frac{du}{\sqrt{2u}} = \frac{2}{\sqrt{2\pi}} \cdot \sqrt{2} \int_0^\infty u^{\frac{1}{2}} e^{-u}du$$
4. Apply gamma function: The integral $\int_0^\infty u^{\frac{1}{2}} e^{-u}du$ is $\Gamma\left(\frac{3}{2}\right)$. We know $\Gamma\left(\frac{3}{2}\right) = \frac{\sqrt{\pi}}{2}$ (using the property $\Gamma(s+1) = s\Gamma(s)$ and $\Gamma\left(\frac{1}{2}\right) = \sqrt{\pi}$).
5. Calculate the final result:
   $$\frac{2}{\sqrt{2\pi}} \cdot \sqrt{2} \cdot \frac{\sqrt{\pi}}{2} = 1$$
   Which is the correct variance for the standard normal distribution. For a general normal $N(\mu, \sigma^2)$, the variance is $\sigma^2$, which follows directly from scaling this result.

Key Takeaway #

The mistake came from a variable substitution that didn't properly account for the quadratic exponent in the normal PDF. By adjusting your substitution to map $x^2$ to the linear term in the gamma function's exponent, you'll end up with an integral over $[0, \infty)$ that aligns perfectly with the gamma function's definition.

内容的提问来源于stack exchange，提问作者hippocampus