I'm reading a paper by Gine where a estimator $T_n$ for $\int f(x)^2dx$ was introduced, where $f$ is the true density function. $T_n$ is defined as $$T_n(h)=\frac{2}{n(n-1)h}\sum_{1\leq i<j\leq n}K\bigg(\frac{X_i-X_j}{h}\bigg)$$ for some observations $X_1,...,X_n$ and a bounded kernel $K$. I understand that to show the consistency, we would like to find some bounds for the bias and the variance of $T_n$. But theorem 1 in Gine try to bound $$\mathbb{E}(T_n(h)-\mathbb{E}T_n(h)-\frac{1}{n}\sum Y_i)^2$$ instead of the variance, where $Y_i=2(f(X_i)-\int f(x)^2dx)$. I can't understand why we need this statement and where does $Y_i$ come. Also I can't find any explaination in the paper. Any hint will be very appreciated!
2
- $\begingroup$ Please check your post for typos, because your final expression simplifies to $\sum E[Y_i],$ which seems unlikely--but the paper is behind a paywall and cannot be consulted by most people to check what you really mean. $\endgroup$whuber– whuber ♦2025-04-08 14:23:57 +00:00Commented Apr 8 at 14:23
- 1$\begingroup$ @whuber Thanks for the hint! I have changed the link for the paper. $\endgroup$toki– toki2025-04-08 14:29:03 +00:00Commented Apr 8 at 14:29
Add a comment |
1 Answer
$\begingroup$ $\endgroup$
1 You can track $Y$ back to the Hoeffding decomposition in equation (2). It's part of the first-order term in the Hoeffding decomposition, and it looks like a sort of first-order bias term. It is plausibly small because $$E[f(X)]=\int f(x)\, dF(x)=\int f(x)\, f(x)dx=\int f^2(x)\, dx$$
- $\begingroup$ Sorry but $2U_n^{^(1)}(\pi_1 R)=2(ER(x,X_1)-ER(X_1.X_2))$. I can't see where $Y_i$ comes from by leting $R=\frac{K(x-y)}{h}$. Actually can we derive consistency from the statement with $Yi$? On page 51 bottom the author already have $Tn-ETn=U_n(R)$, I can't understand why not bound this term directly but need to minus a $Yi$. $\endgroup$toki– toki2025-04-09 08:21:11 +00:00Commented Apr 9 at 8:21