As in the proof of previous theorem, we start with
$\begin{array}{c}\mathrm{\Psi}({Q}_{n}^{\ast})-\mathrm{\Psi}({Q}_{0})=({P}_{n}-{P}_{0}){D}^{\ast}(Q,g)+{P}_{0}({\stackrel{\u02c9}{Q}}_{n}^{\ast}-{\stackrel{\u02c9}{Q}}_{0})\frac{{\stackrel{\u02c9}{g}}_{n}^{\ast}-{\stackrel{\u02c9}{g}}_{0}}{{\stackrel{\u02c9}{g}}_{n}^{\ast}}+{o}_{P}(1/\sqrt{n}),\end{array}$

where we use that ${D}^{\ast}({Q}_{n}^{\ast},{g}_{n}^{\ast})$ falls in a Donsker class with probability tending to 1, and ${P}_{0}\{{D}^{\ast}({Q}_{n}^{\ast},{g}_{n}^{\ast})-{D}^{\ast}(Q,g){\}}^{2}\phantom{\rule{thinmathspace}{0ex}}\mapsto 0$ in probability as $n\text{\hspace{0.17em}}\mapsto \mathrm{\infty}$. The first term yields the first component ${D}^{\ast}(Q,g)$ of the influence curve of ${\mathrm{\psi}}_{n}^{\ast}$.

As in the proof of previous theorem, we decompose this second term as follows:
$\begin{array}{c}{P}_{0}({\stackrel{\u02c9}{Q}}_{n}^{\ast}-{\stackrel{\u02c9}{Q}}_{0})\frac{{\stackrel{\u02c9}{g}}_{n}^{\ast}-{\stackrel{\u02c9}{g}}_{0}}{{\stackrel{\u02c9}{g}}_{n}^{\ast}}={P}_{0}({\stackrel{\u02c9}{Q}}_{n}^{\ast}-\stackrel{\u02c9}{Q}+\stackrel{\u02c9}{Q}-{\stackrel{\u02c9}{Q}}_{0})\frac{{\stackrel{\u02c9}{g}}_{n}^{\ast}-\stackrel{\u02c9}{g}+\stackrel{\u02c9}{g}-{\stackrel{\u02c9}{g}}_{0}}{{\stackrel{\u02c9}{g}}_{n}^{\ast}}\\ ={P}_{0}({\stackrel{\u02c9}{Q}}_{n}^{\ast}-\stackrel{\u02c9}{Q})\frac{{\stackrel{\u02c9}{g}}_{n}^{\ast}-\stackrel{\u02c9}{g}}{{\stackrel{\u02c9}{g}}_{n}^{\ast}}+{P}_{0}({\stackrel{\u02c9}{Q}}_{n}^{\ast}-\stackrel{\u02c9}{Q})\frac{\stackrel{\u02c9}{g}-{\stackrel{\u02c9}{g}}_{0}}{{\stackrel{\u02c9}{g}}_{n}^{\ast}}\\ +{P}_{0}(\stackrel{\u02c9}{Q}-{\stackrel{\u02c9}{Q}}_{0})\frac{{\stackrel{\u02c9}{g}}_{n}^{\ast}-\stackrel{\u02c9}{g}}{{\stackrel{\u02c9}{g}}_{n}^{\ast}}+{P}_{0}(\stackrel{\u02c9}{Q}-{\stackrel{\u02c9}{Q}}_{0})\frac{\stackrel{\u02c9}{g}-{\stackrel{\u02c9}{g}}_{0}}{{\stackrel{\u02c9}{g}}_{n}^{\ast}},\end{array}$

resulting in four terms, which we will denote with Terms 1–4. We will now analyze these four terms.

**Term 1**: The first term ${P}_{0}({\stackrel{\u02c9}{Q}}_{n}^{\ast}-\stackrel{\u02c9}{Q})\frac{{\stackrel{\u02c9}{g}}_{n}^{\ast}-\stackrel{\u02c9}{g}}{\stackrel{\u02c9}{g}}={o}_{P}(1/\sqrt{n})$, by assumption.

**Term 4**: Due to our assumption that ${P}_{0}(\stackrel{\u02c9}{Q}-{\stackrel{\u02c9}{Q}}_{0})(\stackrel{\u02c9}{g}-{\stackrel{\u02c9}{g}}_{0})/\stackrel{\u02c9}{g}=0$ this last term equals:
$\begin{array}{c}-{P}_{0}(\stackrel{\u02c9}{Q}-{\stackrel{\u02c9}{Q}}_{0})(\stackrel{\u02c9}{g}-{\stackrel{\u02c9}{g}}_{0})\frac{({\stackrel{\u02c9}{g}}_{n}^{\ast}-\stackrel{\u02c9}{g})}{{\stackrel{\u02c9}{g}}_{n}^{\ast}\stackrel{\u02c9}{g}}\\ =-{P}_{0}(\stackrel{\u02c9}{Q}-{\stackrel{\u02c9}{Q}}_{0})(\stackrel{\u02c9}{g}-{\stackrel{\u02c9}{g}}_{0})\frac{({\stackrel{\u02c9}{g}}_{n}^{\ast}-\stackrel{\u02c9}{g})}{{\stackrel{\u02c9}{g}}^{2}}+{R}_{{1}_{},n},\end{array}$

where, by assumption,
${R}_{1,n}=-{P}_{0}(\stackrel{\u02c9}{Q}-{\stackrel{\u02c9}{Q}}_{0})(\stackrel{\u02c9}{g}-{\stackrel{\u02c9}{g}}_{0})\frac{({\stackrel{\u02c9}{g}}_{n}^{\ast}-\stackrel{\u02c9}{g}{)}^{2}}{{\stackrel{\u02c9}{g}}^{2}{\stackrel{\u02c9}{g}}_{n}^{\ast}}={o}_{P}(1/\sqrt{n}).$

We proceed as follows:
$\begin{array}{rl}\phantom{\rule{thinmathspace}{0ex}}& -{P}_{0}(\stackrel{\u02c9}{Q}-{\stackrel{\u02c9}{Q}}_{0})(\stackrel{\u02c9}{g}-{\stackrel{\u02c9}{g}}_{0})\frac{({\stackrel{\u02c9}{g}}_{n}^{\ast}-\stackrel{\u02c9}{g})}{{\stackrel{\u02c9}{g}}^{2}}\\ \phantom{\rule{thinmathspace}{0ex}}& =-{P}_{0}(\stackrel{\u02c9}{Q}-{\stackrel{\u02c9}{Q}}_{0})(\stackrel{\u02c9}{g}-A)\frac{({\stackrel{\u02c9}{g}}_{n}^{\ast}-\stackrel{\u02c9}{g})}{{\stackrel{\u02c9}{g}}^{2}}\\ \phantom{\rule{thinmathspace}{0ex}}& =-{P}_{0}\frac{(\stackrel{\u02c9}{Q}-{\stackrel{\u02c9}{Q}}_{0})}{\stackrel{\u02c9}{g}}({\stackrel{\u02c9}{g}}_{n}^{\ast}-\stackrel{\u02c9}{g})+{P}_{0}(\stackrel{\u02c9}{Q}-{\stackrel{\u02c9}{Q}}_{0})\frac{A}{{\stackrel{\u02c9}{g}}^{2}}({\stackrel{\u02c9}{g}}_{n}^{\ast}-\stackrel{\u02c9}{g}).\end{array}$

The first term is asymptotically equivalent with minus Term 3, which shows that Term 3 is canceled out by a component of Term 4 up till a second-order term that is ${o}_{P}(1/\sqrt{n})$, by assumption. The second term equals
$\begin{array}{c}{P}_{0}(\stackrel{\u02c9}{Q}-{\stackrel{\u02c9}{Q}}_{0})\frac{A}{{\stackrel{\u02c9}{g}}^{2}}({\stackrel{\u02c9}{g}}_{n}^{\ast}-\stackrel{\u02c9}{g})=-{P}_{0}(Y-\stackrel{\u02c9}{Q})\frac{A}{{\stackrel{\u02c9}{g}}^{2}}({\stackrel{\u02c9}{g}}_{n}^{\ast}-\stackrel{\u02c9}{g})\\ =-{P}_{0}{E}_{{P}_{0}}(Y-\stackrel{\u02c9}{Q}|A=1,\stackrel{\u02c9}{g},{\stackrel{\u02c9}{g}}_{n}^{\ast})\frac{A}{{\stackrel{\u02c9}{g}}^{2}}({\stackrel{\u02c9}{g}}_{n}^{\ast}-\stackrel{\u02c9}{g})\\ =-{P}_{0}{E}_{{P}_{0}}(Y-\stackrel{\u02c9}{Q}|A=1,\stackrel{\u02c9}{g},{\stackrel{\u02c9}{g}}_{n}^{\ast})\frac{{\stackrel{\u02c9}{g}}_{0,n}^{r}}{{\stackrel{\u02c9}{g}}^{2}}({\stackrel{\u02c9}{g}}_{n}^{\ast}-\stackrel{\u02c9}{g})\\ =-{P}_{0}{E}_{{P}_{0}}(Y-\stackrel{\u02c9}{Q}|A=1,\stackrel{\u02c9}{g})\frac{{\stackrel{\u02c9}{g}}_{0}^{r}}{{\stackrel{\u02c9}{g}}^{2}}({\stackrel{\u02c9}{g}}_{n}^{\ast}-\stackrel{\u02c9}{g})-{P}_{0}({H}_{1}({\stackrel{\u02c9}{g}}_{n}^{\ast})-{H}_{1}(\stackrel{\u02c9}{g}))({\stackrel{\u02c9}{g}}_{n}^{\ast}-\stackrel{\u02c9}{g}),\end{array}$

where ${H}_{1}({\stackrel{\u02c9}{g}}_{n}^{\ast})\equiv {E}_{{P}_{0}}(Y-\stackrel{\u02c9}{Q}|A=1,\stackrel{\u02c9}{g},{\stackrel{\u02c9}{g}}_{n}^{\ast})\frac{{\stackrel{\u02c9}{g}}_{0,n}^{r}}{{\stackrel{\u02c9}{g}}^{2}}$ approximates ${H}_{1}(\stackrel{\u02c9}{g})={E}_{{P}_{0}}(Y-\stackrel{\u02c9}{Q}|A=1,\stackrel{\u02c9}{g})\frac{{\stackrel{\u02c9}{g}}_{0}^{r}}{{\stackrel{\u02c9}{g}}^{2}}$, ${\stackrel{\u02c9}{g}}_{0,n}^{r}={E}_{{P}_{0}}(A|\stackrel{\u02c9}{g},{\stackrel{\u02c9}{g}}_{n}^{\ast})$, and ${\stackrel{\u02c9}{g}}_{0}^{r}={E}_{{P}_{0}}(A|\stackrel{\u02c9}{g})$. Let ${\stackrel{\u02c9}{Q}}_{0,n}^{r}={E}_{{P}_{0}}(Y-\stackrel{\u02c9}{Q}|A=1,\stackrel{\u02c9}{g},{\stackrel{\u02c9}{g}}_{n}^{\ast})$ and ${\stackrel{\u02c9}{Q}}_{0}^{r}={E}_{{P}_{0}}(Y-\stackrel{\u02c9}{Q}|A=1,\stackrel{\u02c9}{g})$. We assumed $\parallel {\stackrel{\u02c9}{g}}_{0,n}^{r}-{\stackrel{\u02c9}{g}}_{0}^{r}{\parallel}_{0}\parallel {\stackrel{\u02c9}{g}}_{n}^{\ast}-\stackrel{\u02c9}{g}{\parallel}_{0}={o}_{P}(1/\sqrt{n})$, and $\parallel {\stackrel{\u02c9}{Q}}_{0,n}^{r}-{\stackrel{\u02c9}{Q}}_{0}^{r}{\parallel}_{0}\parallel {\stackrel{\u02c9}{g}}_{n}^{\ast}-\stackrel{\u02c9}{g}{\parallel}_{0}={o}_{P}(1/\sqrt{n})$, which implies that
${R}_{2,n}={P}_{0}({H}_{1}({\stackrel{\u02c9}{g}}_{n}^{\ast})-{H}_{1}(\stackrel{\u02c9}{g}))({\stackrel{\u02c9}{g}}_{n}^{\ast}-\stackrel{\u02c9}{g})={o}_{P}(1/\sqrt{n}).$

By assumption, ${E}_{0}(A|{W}_{1})$ for some ${W}_{1}$ that is a function of *W*. Therefore, ${\stackrel{\u02c9}{g}}_{0}^{r}(W)={E}_{{P}_{0}}(A|\stackrel{\u02c9}{g}(W))={E}_{{P}_{0}}({E}_{{P}_{0}}(A|{W}_{1})|\stackrel{\u02c9}{g}(W))=\stackrel{\u02c9}{g}(W)$. Thus, it remains to analyze
$-{P}_{0}\frac{{E}_{{P}_{0}}(Y-\stackrel{\u02c9}{Q}|A=1,\stackrel{\u02c9}{g})}{\stackrel{\u02c9}{g}}({\stackrel{\u02c9}{g}}_{n}^{\ast}-\stackrel{\u02c9}{g}).$(4)

This term is analyzed below and it is shown that this term equals
$-({P}_{n}-{P}_{0}){D}_{A}({\stackrel{\u02c9}{Q}}_{0}^{r},\stackrel{\u02c9}{g})+{o}_{P}(1/\sqrt{n}).$

To conclude, we have then shown that the fourth term equals the latter expression minus the third term.

We now analyze (4) which can be represented as $-{P}_{0}\frac{{\stackrel{\u02c9}{Q}}_{0}^{r}}{\stackrel{\u02c9}{g}}({\stackrel{\u02c9}{g}}_{n}^{\ast}-\stackrel{\u02c9}{g})$, where ${\stackrel{\u02c9}{Q}}_{0}^{r}={E}_{{P}_{0}}(Y-\stackrel{\u02c9}{Q}|A=1,\stackrel{\u02c9}{g})$. In this proof, we will use the notation ${H}_{0}^{r}={\stackrel{\u02c9}{Q}}_{0}^{r}/\stackrel{\u02c9}{g}$, ${H}_{n}^{r\ast}={\stackrel{\u02c9}{Q}}_{n}^{r\ast}/{\stackrel{\u02c9}{g}}_{n}^{\ast}$. Since $\stackrel{\u02c9}{g}(W)={E}_{0}(A|{W}_{1})$ for some ${W}_{1}$, and ${\stackrel{\u02c9}{Q}}_{0}^{r}$ is thus also a function of ${W}_{1}$, we have
${P}_{0}\frac{{\stackrel{\u02c9}{Q}}_{0}^{r}}{\stackrel{\u02c9}{g}}\stackrel{\u02c9}{g}={P}_{0}\frac{{\stackrel{\u02c9}{Q}}_{0}^{r}}{\stackrel{\u02c9}{g}}A.$

We now proceed as follows:
$\begin{array}{c}-{P}_{0}{H}_{0}^{r}({\stackrel{\u02c9}{g}}_{n}^{\ast}-\stackrel{\u02c9}{g})=-{P}_{0}{H}_{0}^{r}({\stackrel{\u02c9}{g}}_{n}^{\ast}-A)\\ =-{P}_{0}{H}_{n}^{r\ast}({\stackrel{\u02c9}{g}}_{n}^{\ast}-A)-{P}_{0}({H}_{0}^{r}-{H}_{n}^{r\ast})({\stackrel{\u02c9}{g}}_{n}^{\ast}-A)\\ ={P}_{0}{H}_{n}^{r\ast}(A-{\stackrel{\u02c9}{g}}_{n}^{\ast})-{P}_{0}({H}_{0}^{r}-{H}_{n}^{r\ast})({\stackrel{\u02c9}{g}}_{n}^{\ast}-{E}_{0}(A|\stackrel{\u02c9}{g},{\stackrel{\u02c9}{Q}}_{n}^{r\ast},{\stackrel{\u02c9}{g}}_{n}^{\ast})).\end{array}$

For the second term ${R}_{4,n}$, we can substitute
${\stackrel{\u02c9}{g}}_{n}^{\ast}-{E}_{0}(A|\stackrel{\u02c9}{g},{\stackrel{\u02c9}{Q}}_{n}^{r\ast},{\stackrel{\u02c9}{g}}_{n}^{\ast})=({\stackrel{\u02c9}{g}}_{n}^{\ast}-\stackrel{\u02c9}{g})+{E}_{0}(A|\stackrel{\u02c9}{g},{\stackrel{\u02c9}{Q}}_{0}^{r})-{E}_{0}(A|\stackrel{\u02c9}{g},{\stackrel{\u02c9}{Q}}_{n}^{r\ast},{\stackrel{\u02c9}{g}}_{n}^{\ast}),$

by noting that $\stackrel{\u02c9}{g}={E}_{0}(A|\stackrel{\u02c9}{g},{\stackrel{\u02c9}{Q}}_{0}^{r})$. Thus, this second term results in two terms, one that can be bounded by $\parallel {H}_{n}^{r\ast}-{H}_{0}^{r}{\parallel}_{0}\parallel {\stackrel{\u02c9}{g}}_{n}^{\ast}-\stackrel{\u02c9}{g}{\parallel}_{0}$ and the other is bounded by
$\parallel {H}_{n}^{r\ast}-{H}_{0}^{r}{\parallel}_{0}\parallel {E}_{0}(A|\stackrel{\u02c9}{g},{\stackrel{\u02c9}{Q}}_{0}^{r})-{E}_{0}(A|\stackrel{\u02c9}{g},{\stackrel{\u02c9}{Q}}_{n}^{r\ast},{\stackrel{\u02c9}{g}}_{n}^{\ast}){\parallel}_{0}.$

By assumption, both terms are ${o}_{P}(1/\sqrt{n})$ and thus ${R}_{4,n}={o}_{P}(1/\sqrt{n})$.

Since, by construction of ${g}_{n}^{\ast}$, ${P}_{n}{H}_{n}^{r\ast}(A-{\stackrel{\u02c9}{g}}_{n}^{\ast})=0$, the first term can be written as follows:
$\begin{array}{c}{P}_{0}{H}_{n}^{r\ast}(A-{\stackrel{\u02c9}{g}}_{n}^{\ast})\\ =-({P}_{n}-{P}_{0}){H}_{n}^{r\ast}(A-{\stackrel{\u02c9}{g}}_{n}^{\ast})\\ =-({P}_{n}-{P}_{0}){H}_{0}^{r}(A-\stackrel{\u02c9}{g})+{R}_{5,n},\end{array}$

where ${R}_{5,n}={o}_{P}(1/\sqrt{n})$ if ${P}_{0}\{{D}_{A}({\stackrel{\u02c9}{Q}}_{n}^{r\ast},{\stackrel{\u02c9}{g}}_{n}^{\ast})-{D}_{A}({\stackrel{\u02c9}{Q}}_{0}^{r},\stackrel{\u02c9}{g}){\}}^{2}={o}_{P}(1)$ and ${D}_{A}({\stackrel{\u02c9}{Q}}_{n}^{r\ast},{\stackrel{\u02c9}{g}}_{n}^{\ast})$ falls in a Donsker class with probability tending to 1, and we are reminded that ${D}_{A}(\stackrel{\u02c9}{g},{\stackrel{\u02c9}{Q}}_{0}^{r})={H}_{0}^{r}(A-\stackrel{\u02c9}{g})$. This completes the proof for the fourth term.

**Term 3**: Our analysis of Term 4 showed that Term 3 cancels out and thus that the sum of the third and fourth terms equals $-({P}_{n}-{P}_{0}){D}_{A}({\stackrel{\u02c9}{Q}}_{0}^{r},\stackrel{\u02c9}{g})+{o}_{P}(1/\sqrt{n})$, which yields the second component $-{D}_{A}({\stackrel{\u02c9}{Q}}_{0}^{r},\stackrel{\u02c9}{g})$ of the influence curve of ${\mathrm{\psi}}_{n}^{\ast}$.

**Analysis of Term 2**: Up till a second-order term that can be bounded by $\parallel {\stackrel{\u02c9}{g}}_{n}^{\ast}-\stackrel{\u02c9}{g}{\parallel}_{0}\parallel {\stackrel{\u02c9}{Q}}_{n}^{\ast}-\stackrel{\u02c9}{Q}{\parallel}_{0}={o}_{P}(1/\sqrt{n}$, we can represent Term 2 as
$\begin{array}{c}\left\{{\mathrm{\Psi}}_{2,\stackrel{\u02c9}{g},{\stackrel{\u02c9}{g}}_{0}}({\stackrel{\u02c9}{Q}}_{n}^{\ast})-{\mathrm{\Psi}}_{2,\stackrel{\u02c9}{g},{\stackrel{\u02c9}{g}}_{0}}(\stackrel{\u02c9}{Q})\right\}.\end{array}$

where
$\begin{array}{c}{\mathrm{\Psi}}_{2,\stackrel{\u02c9}{g},{\stackrel{\u02c9}{g}}_{0}}({\stackrel{\u02c9}{Q}}_{n}^{\ast})={P}_{0}{\stackrel{\u02c9}{Q}}_{n}^{\ast}\frac{\stackrel{\u02c9}{g}-{\stackrel{\u02c9}{g}}_{0}}{\stackrel{\u02c9}{g}}.\end{array}$

We have
$\begin{array}{c}{\mathrm{\Psi}}_{2}({\stackrel{\u02c9}{Q}}_{n}^{\ast})-{\mathrm{\Psi}}_{2}(\stackrel{\u02c9}{Q})={P}_{0}\frac{\stackrel{\u02c9}{g}-{\stackrel{\u02c9}{g}}_{0}}{\stackrel{\u02c9}{g}}({\stackrel{\u02c9}{Q}}_{n}^{\ast}-\stackrel{\u02c9}{Q})\\ =-{P}_{0}\frac{A-\stackrel{\u02c9}{g}}{\stackrel{\u02c9}{g}}({\stackrel{\u02c9}{Q}}_{n}^{\ast}-\stackrel{\u02c9}{Q})\\ =-{P}_{0}\frac{{E}_{{P}_{0}}(A|\stackrel{\u02c9}{g},{\stackrel{\u02c9}{Q}}_{n}^{\ast},\stackrel{\u02c9}{Q})-\stackrel{\u02c9}{g}}{\stackrel{\u02c9}{g}}({\stackrel{\u02c9}{Q}}_{n}^{\ast}-\stackrel{\u02c9}{Q}).\end{array}$

Recall that, by our assumption, $\stackrel{\u02c9}{g}={E}_{{P}_{0}}(A|\stackrel{\u02c9}{g},\stackrel{\u02c9}{Q})$. Let ${\stackrel{\u02c9}{g}}_{0,n}^{r}={E}_{{P}_{0}}(A|\stackrel{\u02c9}{g},{\stackrel{\u02c9}{Q}}_{n}^{\ast},\stackrel{\u02c9}{Q})$. By our assumptions,
${P}_{0}\frac{{\stackrel{\u02c9}{g}}_{0,n}^{r}-\stackrel{\u02c9}{g}}{\stackrel{\u02c9}{g}}({\stackrel{\u02c9}{Q}}_{n}^{\ast}-\stackrel{\u02c9}{Q})={o}_{P}(1/\sqrt{n}).$(5)

This proves that ${\mathrm{\Psi}}_{2}({\stackrel{\u02c9}{Q}}_{n}^{\ast})-{\mathrm{\Psi}}_{2}(\stackrel{\u02c9}{Q})={o}_{P}(1/\sqrt{n})$. □

**Remark: Proof of additional result** In this analysis of Term 2, we assumed $\stackrel{\u02c9}{g}={E}_{{P}_{0}}(A|\stackrel{\u02c9}{g},\stackrel{\u02c9}{Q})$, and condition (5). Let us now try to provide a different type of analysis for this Term 2, relying on different conditions. We have
$\begin{array}{rl}{\mathrm{\Psi}}_{2}({\stackrel{\u02c9}{Q}}_{n}^{\ast})-{\mathrm{\Psi}}_{2}(\stackrel{\u02c9}{Q})=\phantom{\rule{thinmathspace}{0ex}}& {P}_{0}\frac{\stackrel{\u02c9}{g}-{\stackrel{\u02c9}{g}}_{0}}{\stackrel{\u02c9}{g}}({\stackrel{\u02c9}{Q}}_{n}^{\ast}-\stackrel{\u02c9}{Q})\\ =\phantom{\rule{thinmathspace}{0ex}}& {P}_{0}\frac{A-\stackrel{\u02c9}{g}}{\stackrel{\u02c9}{g}}(\stackrel{\u02c9}{Q}-{\stackrel{\u02c9}{Q}}_{n}^{\ast})\\ =\phantom{\rule{thinmathspace}{0ex}}& {P}_{0}\frac{{\stackrel{\u02c9}{g}}_{0,n}^{r}-\stackrel{\u02c9}{g}}{{\stackrel{\u02c9}{g}}_{0,n}^{r}\stackrel{\u02c9}{g}}A(Y-{\stackrel{\u02c9}{Q}}_{n}^{\ast}),\end{array}$

where ${\stackrel{\u02c9}{g}}_{0,n}^{r}={E}_{0}(A|\stackrel{\u02c9}{g},\stackrel{\u02c9}{Q},{\stackrel{\u02c9}{Q}}_{n}^{\ast})$, and if we assume that ${P}_{0}\frac{{\stackrel{\u02c9}{g}}_{0,n}^{r}-\stackrel{\u02c9}{g}}{{\stackrel{\u02c9}{g}}_{0,n}^{r}\stackrel{\u02c9}{g}}A(Y-\stackrel{\u02c9}{Q})=0$. The latter equality holds if we target in the TMLE algorithm ${\stackrel{\u02c9}{Q}}_{n}^{\ast}$ with clever covariate ${H}_{Y}({\stackrel{\u02c9}{g}}_{n}^{r\ast},{\stackrel{\u02c9}{g}}_{n}^{\ast})=({\stackrel{\u02c9}{g}}_{n}^{r\ast}-{\stackrel{\u02c9}{g}}_{n}^{\ast})/({\stackrel{\u02c9}{g}}_{n}^{r\ast}{\stackrel{\u02c9}{g}}_{n}^{\ast})A$, where ${\stackrel{\u02c9}{g}}_{n}^{r\ast}$ estimates a non-parametric regression of *A* on ${\stackrel{\u02c9}{Q}}_{n}^{\ast},{\stackrel{\u02c9}{g}}_{n}^{\ast}$, exactly as in Theorem 3. Under that assumption one can now show that we obtain another influence curve component ${D}_{Y}$ defined by
${D}_{Y}(\stackrel{\u02c9}{Q},{\stackrel{\u02c9}{g}}_{0}^{r},\stackrel{\u02c9}{g})=\frac{{\stackrel{\u02c9}{g}}_{0}^{r}-\stackrel{\u02c9}{g}}{{\stackrel{\u02c9}{g}}_{0}^{r}\stackrel{\u02c9}{g}}A(Y-\stackrel{\u02c9}{Q}),$

where ${\stackrel{\u02c9}{g}}_{0}^{r}={E}_{0}(A|\stackrel{\u02c9}{g},\stackrel{\u02c9}{Q})$. Thus, now we have that $\mathrm{\Psi}({Q}_{n}^{\ast})$ is asymptotically linear with influence curve ${D}^{\ast}(Q,g)-{D}_{A}({\stackrel{\u02c9}{Q}}_{0}^{r},\stackrel{\u02c9}{g})-{D}_{Y}(\stackrel{\u02c9}{Q},{\stackrel{\u02c9}{g}}_{0}^{r},\stackrel{\u02c9}{g})$. However, note that if ${\stackrel{\u02c9}{g}}_{0}^{r}=\stackrel{\u02c9}{g}$, i.e. if $\stackrel{\u02c9}{g}={E}_{0}(A|\stackrel{\u02c9}{g},\stackrel{\u02c9}{Q})$, then ${D}_{Y}=0$. To conclude, one can remove the condition that ${\stackrel{\u02c9}{g}}_{n}^{\ast}$ needs to non-parametrically adjust for ${\stackrel{\u02c9}{Q}}_{n}^{\ast}$ as arranged by the TMLE algorithm in Theorem 4 by adding the additional clever covariate ${H}_{Y}({\stackrel{\u02c9}{g}}_{n}^{r\ast},{\stackrel{\u02c9}{g}}_{n}^{\ast})$ to the submodel for ${\stackrel{\u02c9}{Q}}_{n}^{\ast}$ in the TMLE algorithm, and the influence curve will now have another component ${D}_{Y}({P}_{0})$, as in Theorem 3. This results in a generalization of Theorem 3 which does not require that either ${Q}_{n}^{\ast}$ or ${g}_{n}^{\ast}$ is consistent, but only requires that their limits $\stackrel{\u02c9}{g},\stackrel{\u02c9}{Q}$ satisfy ${P}_{0}(\stackrel{\u02c9}{g}-{\stackrel{\u02c9}{g}}_{0})/\stackrel{\u02c9}{g}(\stackrel{\u02c9}{Q}-{\stackrel{\u02c9}{Q}}_{0})=0$. Thus, this latter generalization of Theorem 3 would provide an appropriate theorem for a C-TMLE that does not enforce the non-parametric adjustment for ${\stackrel{\u02c9}{Q}}_{n}^{\ast}$, but still needs to adjust for ${\stackrel{\u02c9}{Q}}_{n}^{\ast}-{\stackrel{\u02c9}{Q}}_{0}$.

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