标准正态分布下累加和超过r的期望抽取次数求解及收敛性疑问

Hi there, let's tackle this problem clearly, since it's a great question that differs notably from the uniform distribution case you mentioned.

First, let's address your core doubt: whether the expected number of draws is defined—this depends entirely on the mean μ of your normal distribution:

• If μ > 0: By the Strong Law of Large Numbers, the cumulative sum will almost surely tend to +∞ as we keep drawing samples. So we will eventually exceed any finite r, and the expected number of draws is finite (well-defined).
• If μ = 0: This becomes a symmetric random walk in continuous space. While the probability of eventually exceeding r is 1, the expected number of draws required to do so is infinite. In this case, you can consider the expectation "undefined" in the sense of being unbounded.
• If μ < 0: The cumulative sum will almost surely tend to -∞, so if r is positive, we will never exceed it. The expectation is completely undefined here (since the process never terminates).

Now, onto the formal modeling. Let's define E(x) as the expected number of additional draws needed when our current cumulative sum is x. Our goal is to find E(0) (starting from a sum of 0).

The recursive equation for E(x) is:

E(x) = 1 + ∫_{-∞}^{∞} E(x + y) f(y) dy, for x ≤ r
E(x) = 0, for x > r

where f(y) is the probability density function of (Y \sim N(\mu, \sigma^2)). As you suspected, the integral here is a convolution of E(·) and f(·). Unfortunately, this integral equation doesn't have a nice closed-form solution for general r, μ, and σ. That's a key difference from the uniform U(0,1) case, where the recursion leads to a solvable differential equation.

That said, we can still get useful results or approximations:

• Asymptotic approximation for large r: When r is much larger than σ, the Law of Large Numbers kicks in. The average increase per draw is μ, so the expected number of draws is approximately ( \frac{r}{\mu} ). This gets more accurate as r grows.
• Numerical methods: If you need a precise value for specific parameters, you can use numerical integration to approximate E(x) iteratively, or run Monte Carlo simulations: generate thousands of independent trials, record how many draws it takes to exceed r each time, then take the average of those counts.
• Special cases: For very small r, you can expand the integral equation around x=r to get a local approximation, but this is only useful for narrow scenarios.

If you're open to alternative approaches, you might also look into generating functions or Markov chain theory (treating the cumulative sum as a continuous-state Markov process), but these won't yield simple closed-form solutions either.

备注：内容来源于stack exchange，提问作者Sunny Chaturvedi