几何分布独立观测值的求和概率计算问题咨询

Hey there! Let's work through this step by step, building on what you already nailed in part (a). First, let's recap the key detail about the geometric distribution we're using here: for (X \sim \text{Geo}(p)) (where (X) counts the number of trials until the first success), the probability mass function (PMF) is:
[ P(X = k) = pq^{k-1} ]
for any integer (k \geq 1), where (q = 1-p).

Step 1: Map valid value pairs for (X_1) and (X_2) #

For (X_1 + X_2 = n) (with (n \geq 2)), (X_1) can take values from (1) to (n-1) — because (X_2 = n - X_1) has to be at least 1 (since geometric variables start at 1). That means we need to sum the probabilities of all these valid pairs:
[ P(X_1 + X_2 = n) = \sum_{i=1}^{n-1} P(X_1 = i) \cdot P(X_2 = n - i) ]
The product here comes from the fact that (X_1) and (X_2) are independent observations.

Step 2: Substitute the geometric PMF into the sum #

Let's plug in the PMF for each term in the sum:

• (P(X_1 = i) = pq^{i-1})
• (P(X_2 = n - i) = pq^{(n-i)-1} = pq^{n - i - 1})

Multiplying these together simplifies nicely:
[ P(X_1 = i) \cdot P(X_2 = n - i) = pq^{i-1} \cdot pq^{n - i - 1} = p^2 q^{(i-1) + (n - i - 1)} = p^2 q^{n-2} ]
Notice this term is constant for all (i) — the (i) cancels out entirely!

Step 3: Evaluate the sum #

We're adding this constant term ((n-1)) times (since we sum from (i=1) to (i=n-1)). So the total probability is:
[ P(X_1 + X_2 = n) = (n-1) \cdot p^2 q^{n-2} ]

Quick check with part (a) #

Let's test this formula with (n=3) (your part (a) case):
[ (3-1)p^2 q^{3-2} = 2p^2 q ]
Which exactly matches the result you calculated earlier. Perfect, that confirms our formula is correct!

A quick note on your earlier attempt #

You tried calculating (P(X_1 \geq 2)) and ended up with (\frac{p}{1-q}) — that's a tiny misstep. Since (q = 1-p), (1-q = p), so (\frac{p}{1-q} = 1), which doesn't make sense. The correct calculation is:
[ P(X_1 \geq 2) = \sum_{k=2}^{\infty} pq^{k-1} = pq \sum_{k=0}^{\infty} q^k = pq \cdot \frac{1}{1-q} = pq \cdot \frac{1}{p} = q ]
Or more simply, (P(X_1 \geq 2) = 1 - P(X_1 = 1) = 1 - p = q). This wasn't directly needed for part (b), but it's good to clear up that confusion!

内容的提问来源于stack exchange，提问作者Tosh