Taylor's theorem for a composition with $\min:\mathbb R^2\to\mathbb R$ and differentiability Lebesgue almost everywhere

Question

Let

• $f\in C^3(\mathbb R)$ with $f>0$ and $$\int f(x)\:{\rm d}x=1$$
• $g:=\ln f$ (and assume $g'$ is Lipschitz continuous)
• $n\in\mathbb N$, $$s(x,y):=\sum_{i=1}^n\left(g(y_i)-g(x_i)\right)$$ and $$h(x,y):=\min\left(1,e^{s(x,\:y)}\right)$$ for $x,y\in\mathbb R^n$
• $x\in\mathbb R^n$ and $Y$ be a $\mathbb R^n$-valued normally distributed random variable on a probability space $(\Omega,\mathcal A,\operatorname P)$ with mean vector $x$ and covariance matrix $\sigma I_n$ for some $\sigma>0$ ($I_n$ denoting the $n\times n$ identity matrix)

I want to make the following argumentation rigorous: By Taylor's theorem, \begin{equation}\begin{split}h(x,Y)-h(x,(x_1,Y_2,\ldots,Y_n))&=\frac{\partial h}{\partial y_1}(x,(x_1,Y_2,\ldots,Y_n))(Y_1-x_1)\\&+\frac12\frac{\partial^2h}{\partial y_1^2}(x,(Z_1,Y_2,\ldots,Y_n))(Y_1-x_1)^2\end{split}\tag1\end{equation} for some real-valued random variable $Z_1$ with $Z_1\in[\min(x_1,Y_1),\max(x_1,Y_1)]$. Thus, \begin{equation}\begin{split}\left.\operatorname E\left[h(x,(y_1,Y_2,\ldots,Y_n))\right]\right|_{y_1\:=\:Y_1}&=\operatorname E\left[\min\left(1,e^A\right)\right]+g'(x_1)\operatorname E\left[1_{\left\{\:A\:<\:0\:\right\}}e^A\right](Y_1-x_1)\\&+\frac12(g''(Z_1)+\left|g'(Z_1)\right|^2)\left.\operatorname E\left[1_{\left\{\:B\:<\:0\:\right\}}e^B\right]\right|_{z_1\:=\:Z_1}(Y_1-x_1)^2.\end{split}\tag2\end{equation} Above, I wrote $$A:=\sum_{i=2}^n(g(Y_i)-g(x_i))$$ and $$B:=g(z_1)-g(x_1)+\sum_{i=2}^n(g(Y_i)-g(x_i))$$ in order to make the equation more readable (you need to replace them where they occur).

Question 1: There are two issues: The first one is that $(x,y)\mapsto\min(x,y)$ is partially differentiable in both arguments except on the diagonal $\Delta_2:=\left\{(x,y)\in\mathbb R^2:x=y\right\}$. Are we able to conclude the existence of $Z_1$ anyway? Note that $$\frac{\partial h}{\partial y_1}(x,y)=\begin{cases}\displaystyle g'(y_1)e^{s(x,\:y)}&\text{, if }s(x,y)<0\\0&\text{, if }s(x,y)>0\end{cases}\tag3$$ and $$\frac{\partial^2h}{\partial y_1^2}(x,y)=\begin{cases}\displaystyle(g''(y_1)+|g'(y_1)|^2)e^{s(x,\:y)}&\text{, if }s(x,y)<0\\0&\text{, if }s(x,y)>0\end{cases}\tag4$$ for all $y\in\mathbb R^n$.

Question 2: The second issue is the case $s(x,y)=0$. In order for $(3)$ to hold, we need to show that the probability of the corresponding event is $0$ (this seems to be related to the question whether the set on which the occurring function is not differentiable has Lebesgue measure $0$; and it's clear that $\Delta$ has Lebesgue measure $0$). How can we do that?

While it's clear that $h$ is partially differentiable with respect to the second variable except on a countable set, it is not clear to me why $h$ is even twice differentiable with respect to the second variable except on a set (at least) of Lebesgue measure $0$ (see this related question).

EDIT: Please take note of this related question which might yield a solution for question 2.

Answer 1

Obviously in the case that $n=1$, we have $s(x,x)=0$ and so if $f'(x) \neq 0$ then $\frac{\partial h}{\partial y_1}$ doesn't exist at $(x,x)$. So I will assume that $n \geq 2$.

Answer to Question 1. The random variable $Z_1$ will pretty much never exist. As an example, take $f$ to be the pdf of a normal distribution of standard deviation $1$, take $n=2$, and take $(x_1,x_2)=(1,0)$; then on the event $\left\{Y_2 \in (-1,1), f(Y_1)>\frac{f(1)f(0)}{f(Y_2)}\right\}$, we have that the LHS of (1) is strictly positive, but the RHS of (1) is nonpositive when $Z_1$ is replaced by any number between $x_1$ and $Y_1$.

Answer to Question 2. By "the corresponding event", I guess you mean the event $\{ s(x,(x_1,Y_2,\ldots,Y_n)) = 0 \}$? This is the same as the event $$ \left\{ \sum_{i=2}^n g(Y_i) = c \right\} $$ where $c=\sum_{i=2}^n g(x_i)$. If there exist $p_1,\ldots,p_{n-1} \in \mathbb{R}$ such that $\sum_{i=1}^{n-1} p_i = c$ and $g^{-1}(\{p_i\})$ has positive Lebesgue measure for each $i$, then clearly this event has positive probability. Otherwise, this event has zero probability:

Theorem. Let $X_1,\ldots,X_N$ be independent random variables each with law equivalent to the Lebesgue measure, and fix any $c \in \mathbb{R}$. Let $g \colon \mathbb{R} \to \mathbb{R}$ be a measurable function and suppose the event $$ E := \left\{ \sum_{i=1}^N g(X_i) = c \right\} $$ has positive probability. Then there exists $(p_1,\ldots,p_N) \in \mathbb{R}^N$ such that $\sum_{i=1}^N p_i = c$ and $g^{-1}(\{p_i\})$ has positive Lebesgue measure for each $i$.

Proof: The statement is obvious for $N=1$ so assume $N \geq 2$. Let $S:=\{p \in \mathbb{R} : \mathrm{Leb}(g^{-1}(\{p\}))>0 \}$. Obviously $S$ is at most countable. We recursively construct $(p_1,\ldots,p_i) \in S^i$, $i \in \{1,\ldots,N-1\}$, such that the event $$ \left\{ \sum_{j=i+1}^N g(X_j) \ = \ c - \sum_{j=1}^i p_j \right\} $$ has positive probability. For $i=N-1$ we then simply define $p_N=c-\sum_{j=1}^{N-1} p_j$, and we are done.

For $i=1$: Fubini's theorem gives that $$ \mathbb{P}\left( c - \sum_{j=2}^N g(X_j) \in S \right) > 0. $$ So since $S$ is at most countable, there exists $p_1 \in S$ such that $$ \mathbb{P}\left( \sum_{j=2}^N g(X_j) = c - p_1 \right) > 0. $$

Now suppose we have $i \in \{1,\ldots,N-2\}$ and $(p_1,\ldots,p_i) \in S^i$ such that the event $$ \left\{ \sum_{j=i+1}^N g(X_j) \ = \ c - \sum_{j=1}^i p_j \right\} $$ has positive probability. Then Fubini's theorem gives that $$ \mathbb{P}\left( c - \left( \sum_{j=1}^i p_j \ \; + \, \sum_{j=i+2}^N g(X_j) \right) \in S \right) > 0. $$ So since $S$ is at most countable, there exists $p_{i+1} \in S$ such that $$ \mathbb{P}\left( \sum_{j=i+2}^N g(X_j) \ = \ c - \sum_{j=1}^{i+1} p_j \right) > 0. $$ So we are done. $\square$