Adaptive Estimation of a Function from its Exponential Radon Transform in Presence of Noise

Abstract

In this article we propose a locally adaptive strategy for estimating a function from its Exponential Radon Transform (ERT) data, without prior knowledge of the smoothness of functions that are to be estimated. We build a non-parametric kernel type estimator and show that for a class of functions comprising a wide Sobolev regularity scale, our proposed strategy follows the minimax optimal rate up to a $\log {n}$ factor. We also show that there does not exist an optimal adaptive estimator on the Sobolev scale when the pointwise risk is used and in fact the rate achieved by the proposed estimator is the adaptive rate of convergence.

Appendix

Proof of Lemma 2.

(a) Let the distribution function for noise be given by $p_{\epsilon }(u) = \frac {1}{\sqrt {2\pi \sigma ^{2}}} e^{\frac {-u^{2}}{2\sigma ^{2}}}$. For the proof of this part, first consider,

$$ \begin{array}{@{}rcl@{}} E_{1}[Z_{n,i}] \!\!\!&=&\!\! \frac{1}{\sqrt{\log n}} E_{(\theta,s)} \!\!\left[E_{1|(\theta,s)} \left[\log \frac{p_{\epsilon} (Y_{i})}{p_{\epsilon} (Y_{i} - T_{\mu} f_{n,1} (\theta_{i},s_{i}))}\right]\right]\\ &=&\!\! \frac{-1}{\sqrt{\log n}} E_{(\theta,s)} \!\!\left[\int \!\log \frac{p_{\epsilon} (Y_{i} - T_{\mu} f_{n,1} (\theta_{i}, s_{i})\!)}{p_{\epsilon}(Y_{i})} p_{\epsilon} (Y_{i} \!- \!T_{\mu} f_{n,1} (\theta_{i}, s_{i})) dY_{i}\!\right]\\ &\geq&\!\! \frac{-1}{\sqrt{\log n}} E_{\theta,s} (T_{\mu} f_{n,1} (\theta_{i}, s_{1}))^{2}. \end{array} $$

Recall that for $f_{n,1} = Ah^{\beta _{1}-1} \eta ((x-x_{0})/h)$ where $h = \left (\frac {\log n}{n}\right )^{\frac {1}{2\beta +1}}$, similar to equation (18) in Abhishek (2022) we have, ${\int \limits }_{Z} (T_{\mu }f_{n,1}(\theta _{i},s_{i}))^{2} ds d\theta \leq c_{8} h^{2\beta +1}$ where c₈ is a constant that can be made as small as desired by choosing a small enough A. In particular, we will choose A such that $\frac {6(\beta _{N}-\beta _{1})}{(2\beta _{1}+1 )(2\beta _{N}+1)}>c_{8}>0$. We remark here that in deriving the estimate for ${\int \limits }_{Z}(T_{\mu }f_{n,1}(\theta _{i},s_{i}))^{2} ds d\theta$ as above, we assume that the design points satisfy a certain feasibility condition (Korostelëv and Tsybakov 1991, Assumption C): $E_{(\theta ,s)}\left [\sum \limits _{i=1}^{n}g(\theta _{i},s_{i})\right ] \leq C_{3} n\int \limits _{Z}g(\theta ,s)dsd \theta$. Thus

$$ \sum\limits_{i=1}^{n} E_{1}[Z_{n,i}] \geq \frac{-1}{\sqrt{\log n}} n E_{\theta,s} (T_{\mu}f_{n,1}(\theta_{i},s_{1}))^{2} \geq - c_{8} \sqrt{\log n}. $$

Proof of part (b) We want to show that ${\sigma _{n}^{2}} = {\sum }_{i=1}^{n}V_{1}[Z_{n,i}]$ is bounded below. First note that from the ‘law of total variance’ V₁[Z_n,i] ≥ E_(𝜃,s) [V_1|(𝜃,s)[Z_n,i]]. Consider

$$ \begin{array}{@{}rcl@{}} Var_{1|(\theta,s)}[Z_{n,i}] &=& \frac{1}{\log n} Var_{1|(\theta,s)} \left[\log \frac{p_{\epsilon} (Y_{i})}{p_{\epsilon} (Y_{i} - T_{\mu} f_{n,1} (\theta_{i}, s_{i}))}\right]\\ &=& \frac{1}{\log n} \left[E_{1|(\theta,s)} \left[\log^{2} \frac{p_{\epsilon} (Y_{i})}{p_{\epsilon} (Y_{i} - T_{\mu} f_{n,1} (\theta_{i},s_{i}))}\right]\right.\\ && \left. - \left( E_{1|(\theta,s)} \left[\log \frac{p_{\epsilon}(Y_{i})}{p_{\epsilon}(Y_{i} - T_{\mu} f_{n,1} (\theta_{i},s_{i}))}\right]\right)^{2}\right]. \end{array} $$

Recall that noise has been assumed to have a Gaussian distribution $\sim N(0,\sigma ^{2})$ Thus,

$$ \begin{array}{@{}rcl@{}} && E_{1|(\theta,s)} \left[\log^{2} \frac{p_{\epsilon} (Y_{i})}{p_{\epsilon} (Y_{i} - T_{\mu} f_{n,1} (\theta_{i}, s_{i}))} \right] \\ && \quad = \frac{1}{\sqrt{2\pi \sigma^{2}}} \int \log^{2} \exp {\left( \frac{(Y_{i} - T_{\mu}f_{n,1} (\theta_{i}, s_{i}))^{2} - {Y}_{i}^{2}}{2\sigma^{2}}\right)} \!\exp {\!\left( \frac{(Y_{i} - T_{\mu} f_{n,1} (\theta_{i}, s_{i}))^{2}}{2 \sigma^{2}}\right)} dY_{i} \\ && \quad = \frac{1}{4 \sigma^{4} \sqrt{2\pi\sigma^{2}}} \int \left( {T}_{\mu}^{4} (f_{n,1} (\theta_{i}, s_{i}) + 4 {Y}_{i}^{2} {T}_{\mu}^{2} f_{n,1} (\theta_{i}, s_{i})) - 4Y_{i} {T}_{\mu}^{3} f_{n,1} (\theta_{i}, s_{i})\right)\\ && \quad \quad \quad \quad \quad \quad \quad \exp{\left( \frac{(Y_{i} - T_{\mu} f_{n,1} (\theta_{i}, s_{i}))^{2}}{2\sigma^{2}}\right)} dY_{i} \\ && \quad = \frac{1}{{4} \sigma^{4}} \left( {T}_{\mu}^{4} f_{n,1} (\theta_{i}, s_{i}) + {4} \sigma^{2} {T}_{\mu}^{2} f_{n,1} (\theta_{i}, s_{i})\right). \end{array} $$

(A.1)

On the other hand,

$$ \begin{array}{@{}rcl@{}} \left( E_{1|(\theta,s)} \left[\log \frac{p_{\epsilon}(Y_{i})}{p_{\epsilon}(Y_{i}-T_{\mu}f_{n,1}(\theta_{i},s_{i}))}\right]\right)^{2} = \frac{{T}_{\mu}^{4} f_{n,1} (\theta_{i}, s_{i})}{4\sigma^{4}}. \end{array} $$

Thus $Var_{1|(\theta ,s)}[Z_{n,i}]=\frac {{4}(T_{\mu }f_{n,1}(\theta _{i},s_{i}))^{2}}{\log n\sigma ^{2}}$ and hence,

$$ \begin{array}{@{}rcl@{}} \sum\limits_{i=1}^{n} E_{\theta,s} [Var_{1|(\theta,s)}[Z_{n,i}]] &=& \frac{n}{\sigma^{2}\log n} {\int}_{{}Z} (T_{\mu}f_{n,1}(\theta_{i},s_{i}))^{2} dsd \theta\\ &=& c_{10}\frac{n}{\log n}h^{2\beta+1}>0. \end{array} $$

Proof of part (c)

$$ \begin{array}{@{}rcl@{}} \!\!\!\!\!E_{1} \lvert U_{n,i}^{3}\rvert\!\!\! &=&\!\!\! \frac{1}{{\sigma_{n}^{3}}}E_{1}\lvert Z_{n,i}^{3}-\!(E_{1}[Z_{n,i}])^{3} \!-3 (Z_{n,i})^{2}E_{1}[Z_{n,i}]+3(Z_{n,i})(E_{1}[Z_{n,i}])^{2} \rvert\\ &\leq&\!\!\! \frac{1}{{\sigma_{n}^{3}}}[E_{1}\lvert Z_{n,i}\rvert^{3}+(E_{1}\lvert Z_{n,i}\rvert)^{3} +3 E_{1}\lvert Z_{n,i} \rvert^{2}E_{1}\lvert Z_{n,i} \rvert\\&&+3E_{1}\lvert Z_{n,i} \rvert(E_{1}\lvert Z_{n,i} \rvert)^{2} ] \\ &\leq&\!\!\! \frac{1}{{\sigma_{n}^{3}}}[E_{1}\lvert Z_{n,i}\rvert^{3} + (E_{1}\lvert Z_{n,i} \rvert)^{3} + 3 E_{1}\lvert Z_{n,i} \rvert^{2}E_{1}\lvert Z_{n,i} \rvert + (E_{1}\lvert Z_{n,i} \rvert)^{3}].\\ \end{array} $$

(A.2)

Now we consider each of the above terms one by one. First of all $E_{1}\lvert Z_{n,i}\rvert =E_{\theta ,s}[E_{1|(\theta ,s)}\lvert Z_{n,i}\rvert ]$. Thus using Pinsker’s second inequality to calculate:

$$ \begin{array}{@{}rcl@{}} E_{1|(\theta,s)}\lvert Z_{n,i}\rvert &=& \frac{1}{\sqrt{\log n}}\int\left\lvert \log \frac{p_{\epsilon}(Y_{i})}{p_{\epsilon}(Y_{i}-T_{\mu}f_{n,1}(\theta_{i},s_{i}))}\right\rvert p_{\epsilon}(Y_{i}-T_{\mu}f_{n,1}(\theta_{i},s_{i})) dY_{i} \\ &\leq& \frac{1}{\sqrt{\log n}} \left[\frac{T_{\mu}f_{n,1}(\theta_{i},s_{i})}{\sigma}+\frac{{T}_{\mu}^{2}f_{n,1}(\theta_{i},s_{i})}{2\sigma^{2}}\right] \text{[Tsybakov, 2009, Lemma 2.5]}.\\ \end{array} $$

(A.3)

Also note that since the cylinder Z = S¹ × [− 1,1] has finite measure, we have:

$$ \begin{array}{@{}rcl@{}} \left\lvert {\int}_{{}Z} T_{\mu} f_{n,1} (\theta,s) dsd \theta \right\rvert &\leq& {\int}_{{}Z} \lvert T_{\mu} f_{n,1} (\theta,s) \rvert dsd \theta\\ &\leq& c_{10} \left( {\int}_{{}Z} \lvert T_{\mu} f_{n,1} (\theta,s) \rvert^{2} dsd\theta\right)^{1/2}. \end{array} $$

(A.4)

From inequalities (A.3) and (A.4), we get:

$$ \begin{array}{@{}rcl@{}} E_{1}\rvert{Z_{n,i}}\lvert &\leq& \frac{c_{11}}{\sqrt{\log n}} \left[\left( \frac{\log n}{n}\right)^{1/2} + \left( \frac{\log n}{n}\right)\right] \\ &\leq& \frac{c_{12}}{\sqrt{\log n}} \left( \frac{\log n}{n}\right)^{1/2} \quad \left( \!0\!<\frac{\log n}{n}\!\leq \!\left( \frac{\log n}{n}\right)^{1/2}<1 \text{ for } n\geq 3\!\right)\!. \end{array} $$

Finally,

$$ \begin{array}{@{}rcl@{}} \sum\limits_{i=1}^{n} \left( E_{1}\rvert{Z_{n,i}}\lvert\right)^{3} \leq c_{12}n \left( \frac{1}{n}\right)^{3/2} \to 0 \quad \text{as }n\to\infty. \end{array} $$

(A.5)

Using Eq. A.1, we have:

$$ E_{1|(\theta,s)} [\rvert{Z_{n,i}}\lvert^{2}] \leq \frac{1}{{4}\sigma^{4} \log n} \left( T_{\mu}^{4}f_{n,1}(\theta_{i},s_{i}) + {4} \sigma^{2} {T}_{\mu}^{2} f_{n,1} (\theta_{i},s_{i})\right). $$

Then using the fact that $\left \lvert T_{\mu } f_{n,1}(\theta _{i},s_{i})\right \rvert \leq c_{13} h^{\frac {\beta }{2\beta +1}} = c_{13} \left (\frac {\log n}{n}\right )^{\frac {\beta }{2\beta +1}}$,

$$ \begin{array}{@{}rcl@{}} E_{1}[\rvert{Z_{n,i}}\lvert^{2}] &\leq& \frac{c_{14}}{\log n} \left[\left( \frac{\log n}{n}\right)^{\frac{4\beta}{2\beta+1}} + \left( \frac{\log n}{n}\right)^{\frac{2\beta}{2\beta+1}} \right] \\ &\leq& \frac{c_{15}}{\log n} \left[\left( \frac{\log n}{n}\right)^{\frac{2\beta}{2\beta+1}}\right]\\ && \quad \quad \left( 0<\left( \frac{\log n}{n}\right)^{\frac{4\beta}{2\beta+1}} \leq \left( \frac{\log n}{n}\right)^{\frac{2\beta}{2\beta+1}} < 1 \text{ for } n\geq 3\right). \end{array} $$

Finally,

$$ \begin{array}{@{}rcl@{}} \sum\limits_{i=1}^{n} E_{1} \lvert Z_{n,i}\rvert^{2} E_{1} \lvert Z_{n,i} \rvert &\leq& c_{16} \frac{n}{\log n} \frac{1}{\sqrt{\log n}} \left( \frac{\log n}{n}\right)^{\frac{6\beta+1}{4\beta+2}} \\ &\leq& c_{16} \frac{1}{\sqrt{\log n}} \left( \frac{\log n}{n}\right)^{\frac{2\beta-1}{4\beta+2}} \quad \to 0 \text{ as } n \to \infty.\\ \end{array} $$

(A.6)

Now we consider ${\sum }_{i=1}^{n} E_{1}\lvert Z_{n,i}\rvert ^{3}$. For this we first evaluate:

$$ \begin{array}{@{}rcl@{}} &&E_{1|(\theta,s)}\lvert Z_{n,i}\rvert^{3}\\ &=& \frac{1}{(\log n)^{3/2}}\int \left\lvert \log \frac{p_{\xi}(Y_{i})}{p_{\epsilon}(Y_{i} - T_{\mu} f_{n,1} (\theta_{i}, s_{i}))} \right\rvert^{3} p_{\epsilon}(Y_{i}-T_{\mu}f_{n,1}(\theta_{i},s_{i})) dY_{i}\\ &\leq& \frac{1}{(\log n)^{3/2}}\int\left[ \left\lvert {T}_{\mu}^{2} f_{n,1} (\theta_{i}, s_{i}) - 2Y_{i} T_{\mu} f_{n,1} (\theta_{i}, s_{i}) \right\rvert \right]^{3} p_{\epsilon} (Y_{i} - T_{\mu}f_{n,1} (\theta_{i}, s_{i})) dY_{i}\\ &\leq& \frac{1}{(\log n)^{3/2}} \int\left[\vphantom{\frac{1}{2}} \left\lvert T^{6}_{\mu}f_{n,1}(\theta_{i},s_{i}) \right\rvert+8 \left\lvert Y_{i} \right\rvert^{3}\left\lvert T_{\mu}^{3}f_{n,1}(\theta_{i},s_{i})\right\rvert+12\left\lvert Y_{i} \right\rvert^{2}\left\lvert T_{\mu}^{4}f_{n,1}(\theta_{i},s_{i}) \right\rvert \right.\\ &&\left. + 6\left\lvert Y_{i}\right\rvert\left\lvert T_{\mu}^{5}f_{n,1}(\theta_{i},s_{i}) \right\rvert \vphantom{\frac{1}{2}}\right] \frac{\exp{(-(Y_{i}-T_{\mu}f_{n,1}(\theta_{i},s_{i}))^{2}/2\sigma^{2}})}{\sqrt{2\pi\sigma^{2}}} dY_{i}\\ &\leq& \frac{c_{16}}{(\log n)^{3/2}}\left( \frac{\log n}{n}\right)^{\frac{6\beta}{2\beta+1}}, \end{array} $$

where the last inequality follows from the previous one by integrating each term and using the fact that $\lvert T_{\mu } f_{n,1} (\theta _{i}, s_{i}) \rvert \leq c_{13} \left (\frac {\log n}{n}\right )^{\frac {\beta }{2\beta +1}}$. Thus:

$$ \begin{array}{@{}rcl@{}} \sum\limits_{i=1}^{n} E_{1} \lvert Z_{n,i}\rvert^{3} &\leq& \frac{c_{17}}{(\log n)^{1/2}} \frac{n}{\log n} \left( \frac{\log n}{n}\right)^{\frac{6\beta}{2\beta+1}}\\ &\leq& \frac{c_{17}}{(\log n)^{1/2}} \left( \frac{\log n}{n}\right)^{\frac{4\beta-1}{2\beta+1}} \quad \to 0 \text{ as } n \to \infty. \end{array} $$

(A.7)

Equations A.2, A.5, A.6 and A.7 together prove part (c). □