基于给定不等式的算法样本量Big-O符号估算问题

Great question! Let's walk through how to figure out the asymptotic scaling of $n$ using Big-O notation for this self-referential sample size condition. First, let's restate the original inequality to make it easier to work with:

$n \ge \frac{C}{\epsilon^2} \left( \log \frac{1}{\delta} + D \log \frac{n}{D} \right)$

This is a tricky one because $n$ shows up on both sides—linear on the left, inside a logarithm on the right. But we can use standard asymptotic analysis tricks to find its Big-O scaling.

Step 1: Simplify the notation #

Let's define a couple of shorthand variables to clean things up:

• Let $K = \frac{C}{\epsilon^2}$ (this is the core scaling factor tied to the error tolerance $\epsilon$)
• Let $L = \log\frac{1}{\delta}$ (this captures the confidence level, since smaller $\delta$ means higher confidence)

Now the inequality becomes:
$n \ge K \left( L + D \log\left(\frac{n}{D}\right) \right)$

Step 2: Key growth rate observation #

The critical thing here is that $\log(n)$ grows extremely slowly compared to linear terms in $n$. Even though $n$ is in the log on the right, it won't dominate the overall scaling of $n$—it'll only add a logarithmic correction to the main terms we already see.

Step 3: Iterative substitution (the go-to trick for self-referential bounds) #

We can't solve for $n$ directly, so we use iterative substitution to guess and refine our bound:

1. First pass: Assume the $\log(n)$ term is driven by the main known terms. Let's guess $n$ is roughly proportional to $K(L + D \log(KL/D))$. We substitute this rough estimate into the log term on the right.
2. Second pass: Plug this guess back into the right-hand side to check. The log term becomes:
   $\log\left(\frac{n}{D}\right) = \log\left( \frac{K(L + D \log(KL/D))}{D} \right)$
   For large enough values of $L$ (small $\delta$) or $D$ (high dimension), the $D \log(...) $ term inside dominates, so this simplifies to $\log(K) + \log\left(\log\left(\frac{KL}{D}\right)\right)$. This is a log-log term, which grows even slower than a regular log—so it doesn't change the overall asymptotic scaling.

Step 4: Final Big-O bound #

Putting it all together, the required sample size $n$ has the following asymptotic scaling:
$n = O\left( \frac{C}{\epsilon^2} \left( \log\frac{1}{\delta} + D \log\left( \frac{C D}{\epsilon^2 \delta} \right) \right) \right)$

If you prefer a more expanded form, you can break down the log term:
$\log\left( \frac{C D}{\epsilon^2 \delta} \right) = \log(C D) + 2\log\left(\frac{1}{\epsilon}\right) + \log\left(\frac{1}{\delta}\right)$
Which gives us an alternative version of the bound:
$n = O\left( \frac{C}{\epsilon^2} \left( \log\frac{1}{\delta} + D \log\frac{1}{\epsilon} + D \log D \right) \right)$

Quick intuition #

• The $\frac{C}{\epsilon^2}$ term is the standard scaling for many statistical algorithms—sample size grows inversely with the square of your error tolerance $\epsilon$.
• The $\log\frac{1}{\delta}$ term accounts for confidence: smaller $\delta$ (higher confidence) needs more samples.
• The $D \log(...) $ term is the "dimension tax": higher-dimensional problems (larger $D$) need more samples, but only logarithmically in the dimension and other parameters.

内容的提问来源于stack exchange，提问作者eulerup