顺序统计量与随机变量函数：求解Y的分布密度问题求助

Hey there! Let's break this down step by step—since (X \sim Bin(5, 0.2)) is a discrete random variable, we don't need to use Jacobian methods (those are reserved for continuous variable transformations). Instead, we can directly enumerate all possible values of (X), compute the corresponding (Y) values, and group their probabilities to get (Y)'s distribution.

Step 1: List all possible values and probabilities of X #

For (X \sim Bin(n=5, p=0.2)), the possible values are (k = 0,1,2,3,4,5). The probability mass function (PMF) is:
[P(X=k) = \binom{5}{k}(0.2)^k(0.8){5-k}]
Calculating each probability:

• (P(X=0) = 0.32768)
• (P(X=1) = 0.4096)
• (P(X=2) = 0.2048)
• (P(X=3) = 0.0512)
• (P(X=4) = 0.0064)
• (P(X=5) = 0.00032)

Step 2: Compute Y for each X value #

Recall (Y = \frac{1}{2}(Max(X,4) + Min(X,2))). Let's calculate (Y) for each (k):

• When (k=0): (Max(0,4)=4), (Min(0,2)=0) → (Y = \frac{4+0}{2} = 2)
• When (k=1): (Max(1,4)=4), (Min(1,2)=1) → (Y = \frac{4+1}{2} = 2.5)
• When (k=2): (Max(2,4)=4), (Min(2,2)=2) → (Y = \frac{4+2}{2} = 3)
• When (k=3): (Max(3,4)=4), (Min(3,2)=2) → (Y = \frac{4+2}{2} = 3)
• When (k=4): (Max(4,4)=4), (Min(4,2)=2) → (Y = \frac{4+2}{2} = 3)
• When (k=5): (Max(5,4)=5), (Min(5,2)=2) → (Y = \frac{5+2}{2} = 3.5)

Step 3: Group probabilities for each Y value #

Now we can map each unique (Y) value to the sum of probabilities from the corresponding (X) values:

• (P(Y=2) = P(X=0) = 0.32768)
• (P(Y=2.5) = P(X=1) = 0.4096)
• (P(Y=3) = P(X=2) + P(X=3) + P(X=4) = 0.2048 + 0.0512 + 0.0064 = 0.2624)
• (P(Y=3.5) = P(X=5) = 0.00032)

Why Jacobian isn't needed here #

Jacobian transformation is used when converting the probability density function (PDF) of a continuous random variable to another continuous variable. Since (X) is discrete, we only need to track how each discrete value maps to (Y) and aggregate probabilities—no calculus or Jacobian determinants required!

Also, note that (Max(X,4)) and (Min(X,2)) are just piecewise functions for a single random variable, not order statistics (which apply to multiple independent random variables from the same distribution). So we don't need to use order statistic distribution formulas here either—simple enumeration works perfectly.

内容的提问来源于stack exchange，提问作者sergio.azevedo