两个不同成功概率的几何随机变量之和的概率密度求解疑问

Hey Bob, let's walk through the issues in your current solution and get to the correct probability mass function (note: geometric distribution is discrete, so we talk about PMF instead of PDF) for (Z = X + Y):

Key Issues in Your Solution #

• Unclear Geometric Distribution Definition: Geometric distributions have two common formulations, and you didn't specify which one you're using:
  • Version 1: Counts failures before the first success (supports values (0,1,2,...))
  • Version 2: Counts total trials until the first success (supports values (1,2,3,...))
    Your calculation includes (P(Z=0)), which only makes sense for Version 1. If you were using Version 2, (Z) can't be 0 or 1—this ambiguity breaks the foundation of your calculation.
• Incomplete Probability Expression: Your sum term cuts off mid-way ((\sum_{i=0}^{n} p_1(1-...)), so it's impossible to verify if you're using the correct PMF for (X) and (Y). Even if you intended the right form, you didn't simplify the sum to get a closed-form result, which is the goal of this problem.
• Potential Range Errors: Depending on the geometric distribution version, your summation bounds might be wrong. For Version 2, (X) and (Y) can't be 0, so the sum for (P(Z=n)) should start at (i=1) and end at (i=n-1), not (i=0) to (i=n).

Correct Calculation for (Z = X + Y) #

First, let's clarify both common geometric distribution definitions and solve for each:

Case 1: Geometric Distribution (Failures Before First Success)

Here, (X,Y \in {0,1,2,...}), with:
[
P(X=i) = p_1(1-p_1)^i, \quad P(Y=j) = p_2(1-p_2)^j
]
For (Z = n) ((n \geq 0)):
[
P(Z=n) = \sum_{i=0}^{n} P(X=i)P(Y=n-i) = \sum_{i=0}^{n} p_1(1-p_1)^i p_2(1-p_2)^{n-i}
]

• If (p_1 \neq p_2), this simplifies to a closed form using geometric series summation:
  [
  P(Z=n) = \frac{p_1p_2}{p_1 - p_2} \left[ (1-p_2)^{n+1} - (1-p_1)^{n+1} \right]
  ]
• If (p_1 = p_2 = p), the sum becomes:
  [
  P(Z=n) = (n+1)p^2(1-p)n
  ]

Case 2: Geometric Distribution (Trials Until First Success)

Here, (X,Y \in {1,2,3,...}), with:
[
P(X=i) = p_1(1-p_1)^{i-1}, \quad P(Y=j) = p_2(1-p_2)^{j-1}
]
For (Z = n) ((n \geq 2)):
[
P(Z=n) = \sum_{i=1}^{n-1} P(X=i)P(Y=n-i) = \sum_{i=1}^{n-1} p_1(1-p_1)^{i-1} p_2(1-p_2)^{n-i-1}
]

• If (p_1 \neq p_2), simplifying gives:
  [
  P(Z=n) = \frac{p_1p_2}{p_1 - p_2} \left[ (1-p_2)^{n-1} - (1-p_1)^{n-1} \right]
  ]
• If (p_1 = p_2 = p), the sum simplifies to:
  [
  P(Z=n) = (n-1)p^2(1-p){n-2}
  ]

内容的提问来源于stack exchange，提问作者Bob