已知$X\sim N(0,1)$，求$Y=X|X>0$的分布及PDF推导方法

Great question! When dealing with a conditional distribution where we restrict a single variable to a subset of its support, we can use the core definition of conditional probability applied to the cumulative distribution function (CDF), then differentiate to get the probability density function (PDF). Let's walk through this step by step.

Step 1: Recall Key Properties of Standard Normal (X) #

First, let's recap what we know about (X \sim N(0,1)):

• Its PDF is (\phi(x) = \frac{1}{\sqrt{2\pi}} e^{-x2/2}) for all real (x)
• Its CDF is (\Phi(x) = P(X \leq x) = \int_{-\infty}^x \phi(t) dt)
• Due to symmetry around 0, (P(X > 0) = 1/2) and (\Phi(0) = 0.5)

Step 2: Compute the Conditional CDF (F_Y(y)) #

By definition, the CDF of (Y = X \mid X > 0) is:
[ F_Y(y) = P(Y \leq y) = P(X \leq y \mid X > 0) ]

Using the basic conditional probability formula (P(A \mid B) = \frac{P(A \cap B)}{P(B)}), we rewrite this as:
[ F_Y(y) = \frac{P(X \leq y \text{ and } X > 0)}{P(X > 0)} ]

Now we split into two cases based on the value of (y):

Case 1: (y \leq 0) #

If (y \leq 0), there's no overlap between (X \leq y) and (X > 0), so (P(X \leq y \text{ and } X > 0) = 0). Thus:
[ F_Y(y) = \frac{0}{1/2} = 0 ]

Case 2: (y > 0) #

If (y > 0), (X \leq y) and (X >0) simplifies to (0 < X \leq y). The probability of this event is (P(0 < X \leq y) = \Phi(y) - \Phi(0)). Since (\Phi(0)=0.5), substitute that in:
[ F_Y(y) = \frac{\Phi(y) - 0.5}{1/2} = 2\Phi(y) - 1 ]

Putting it all together, the CDF of (Y) is:
[ F_Y(y) = \begin{cases}
0 & \text{if } y \leq 0 \
2\Phi(y) - 1 & \text{if } y > 0
\end{cases} ]

Step 3: Differentiate to Get the Conditional PDF (f_Y(y)) #

The PDF is the derivative of the CDF with respect to (y):

• For (y \leq 0): (F_Y(y) = 0), so (f_Y(y) = 0)
• For (y > 0): We know (\frac{d}{dy}\Phi(y) = \phi(y)), so:
  [ f_Y(y) = \frac{d}{dy}(2\Phi(y) -1) = 2\phi(y) ]

Substituting the standard normal PDF (\phi(y)):
[ f_Y(y) = 2 \cdot \frac{1}{\sqrt{2\pi}} e^{-y2/2} = \sqrt{\frac{2}{\pi}} e^{-y2/2} ]

So the full PDF of (Y) is:
[ f_Y(y) = \begin{cases}
0 & \text{if } y \leq 0 \
\sqrt{\frac{2}{\pi}} e^{-y2/2} & \text{if } y > 0
\end{cases} ]

Step 4: Verify the PDF is Valid #

To confirm this is a valid PDF, check that it integrates to 1 over all real (y):
[ \int_{-\infty}^\infty f_Y(y) dy = \int_0^\infty \sqrt{\frac{2}{\pi}} e^{-y2/2} dy ]
Since (\int_0^\infty \phi(y) dy = 0.5), multiplying by 2 gives:
[ 2 \cdot 0.5 = 1 ]
Which checks out!

This distribution is often called the truncated standard normal distribution (truncated below at 0).

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