含分布函数的分部积分：标准正态积分等式证明问询

Alright, let's tackle this systematically—first I'll explain how integration by parts works when dealing with distribution functions, then we'll apply that framework to prove your normal distribution integral identity.

First, recall the standard integration by parts formula:
$$\int_a^b u , dv = \left. uv \right|_a^b - \int_a^b v , du$$

When working with distribution functions (let's denote a general distribution function as $F(x)$, with corresponding probability density function $f(x) = F'(x)$), we often encounter integrals of the form $\int g(x) f(x) dx$. Since $f(x)dx = dF(x)$, we can rewrite this integral as $\int g(x) dF(x)$, which fits neatly into the integration by parts structure:

• Let $u = g(x)$, so $du = g'(x)dx$
• Let $dv = dF(x)$, so $v = F(x)$

Substituting into the formula gives:
$$\int_a^b g(x) f(x) dx = \left. g(x)F(x) \right|_a^b - \int_a^b F(x) g'(x) dx$$

Key things to remember for boundary terms:

• For most well-behaved distributions (like normal), $\lim_{x \to \infty} F(x) = 1$ and $\lim_{x \to -\infty} F(x) = 0$
• If $g(x)$ doesn't grow faster than the density decays (e.g., linear functions with normal distributions), boundary terms at $\pm\infty$ will vanish (since exponential decay beats polynomial growth)

We need to prove:
$$\text{EE} = \int_{-\mu/\sigma}^{+\infty} (\mu + \sigma x)\phi(x)\ dx = \mu\Phi(\mu/\sigma) + \sigma\phi(\mu/\sigma)$$

First, split the integral into two separate terms—this simplifies things a lot:
$$\text{EE} = \mu \int_{-\mu/\sigma}^\infty \phi(x) dx + \sigma \int_{-\mu/\sigma}^\infty x\phi(x) dx$$

Let's handle each term one by one.

Term 1: The $\mu$ Component #

Recall that $\Phi(x) = \int_{-\infty}^x \phi(t)dt$, so $\int_{a}^\infty \phi(x)dx = 1 - \Phi(a)$. Here, our lower limit is $a = -\mu/\sigma$, so:
$$\mu \int_{-\mu/\sigma}^\infty \phi(x) dx = \mu \left(1 - \Phi\left(-\frac{\mu}{\sigma}\right)\right)$$

Now use the symmetry of the standard normal distribution: $\Phi(-z) = 1 - \Phi(z)$ (since the density is even). Substitute that in:
$$\mu \left(1 - (1 - \Phi\left(\frac{\mu}{\sigma}\right))\right) = \mu\Phi\left(\frac{\mu}{\sigma}\right)$$

That's exactly the first term on the right-hand side of our target identity.

Term 2: The $\sigma$ Component #

Now we need to compute $\sigma \int_{-\mu/\sigma}^\infty x\phi(x) dx$. Here's where integration by parts (or a handy derivative trick) comes in. First, remember the derivative of the standard normal density:
$$\phi'(x) = \frac{d}{dx} \left( \frac{1}{\sqrt{2\pi}} e^{-x2/2} \right) = -x\phi(x)$$

Rearranging that gives $x\phi(x) = -\phi'(x)$. This is a huge shortcut—instead of doing full integration by parts, we can directly integrate using this identity:
$$\int x\phi(x) dx = -\phi(x) + C$$

Now evaluate from $-\mu/\sigma$ to $\infty$:
$$\int_{-\mu/\sigma}^\infty x\phi(x) dx = \left. -\phi(x) \right|_{-\mu/\sigma}^\infty$$

Compute the boundary terms:

• As $x \to \infty$, $\phi(x) \to 0$ (exponential decay dominates), so the upper limit term is $-0 = 0$
• At the lower limit $x = -\mu/\sigma$, we have $-\phi\left(-\frac{\mu}{\sigma}\right)$

Putting it all together:
$$\int_{-\mu/\sigma}^\infty x\phi(x) dx = 0 - \left( -\phi\left(-\frac{\mu}{\sigma}\right) \right) = \phi\left(-\frac{\mu}{\sigma}\right)$$

Again, use the evenness of the normal density: $\phi(-z) = \phi(z)$. So this becomes $\phi\left(\frac{\mu}{\sigma}\right)$. Multiply by $\sigma$:
$$\sigma \int_{-\mu/\sigma}^\infty x\phi(x) dx = \sigma\phi\left(\frac{\mu}{\sigma}\right)$$

Combine Both Terms #

Adding the two results together gives:
$$\text{EE} = \mu\Phi\left(\frac{\mu}{\sigma}\right) + \sigma\phi\left(\frac{\mu}{\sigma}\right)$$

Which matches exactly the identity we needed to prove!

内容的提问来源于stack exchange，提问作者J. Pete