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Theorem 1: In the known exponential bounds for martingales, the conditional variances cannot be replaced by the unconditional ones.

Proof: Otherwise, we would most likely have such bounds. $\Box$ :-)

This "proof" of "Theorem 1" is not so non-serious as it may look.

Perhaps more seriously, we have

Theorem 2: The following statement is false:

There is a real constant $c>0$ such that for all natural $n$, all real $y>0$, all real $B>0$, and all martingale difference sequences $(X_1,\dots,X_n)$ such that \begin{equation*} X_i\le y \ \forall i\quad\text{and}\quad\sum_{i=1}^n Var\,X_i\le B^2\label{0}\tag{0} \end{equation*} we have \begin{equation*} P\Big(\sum_{i=1}^n X_i\ge x\Big)\le\exp\frac{-cx^2}{B^2+xy}\label{1}\tag{1} \end{equation*} for all real $x>0$.

Proof: This proof would be a bit simpler if, instead of using Corollary 2 in the Pinelis--Utev paper, you used the better bound in Theorem 3 in that paper. Indeed, one can show that, at least in the case when the $X_i$'s are conditionally symmetric (given $\mathcal F_{i-1}$), that theorem implies the Rosenthal-type inequality \begin{equation*} ES_n^4\ll B^4+A^{(4)}_n, \end{equation*} where $$S_n:=\sum_{i=1}^n X_i,$$ $a\ll b$ means $a\le Cb$ for some real $C$ depending only on $c$, and \begin{equation*} A^{(p)}_n:=\sum_{i=1}^n E|X_i|^p. \end{equation*}

Because the bound in \eqref{1} is suboptimal, it only implies an ugly version of the Rosenthal-type inequality:

Lemma 1: If the highlighted statement is true, then for conditionally symmetric martingale difference sequences $(X_1,\dots,X_n)$ such that $\sum_{i=1}^n Var\,X_i\le B^2$ we have \begin{equation*} ES_n^4\ll B^4+A^{(6)}_n/B^2. \label{2}\tag{2} \end{equation*}

This lemma will be proved at the end of this answer.

Now consider the following construction of a conditionally symmetric martingale difference sequence $(X_1,\dots,X_n)$: Let $V_1:=R_1$, where $R_1$ is a Rademacher random variable, so that $P(R_1=\pm1)=1/2$. For natural $k\ge2$, let \begin{equation*} V_k:=a_k R_k,\quad a_k:=\frac1{\sqrt{k\ln k}}, \end{equation*} where $R_2,R_3,\dots$ are independent copies of $R_1$. Let then $X_1:=V_1$, and for natural $k\ge2$ let \begin{equation*} X_k:=S_{k-1}V_k, \end{equation*} where $S_j:=\sum_{i=1}^j X_i$, as before. So, for natural $k\ge2$, \begin{equation*} S_k=S_{k-1}(1+V_k). \end{equation*} So, for any even natural $p$ and any natural $k\ge2$, we have $M_k^{(p)}:=ES_k^p=M_{k-1}^{(p)} E(1+V_k)^p$ and hence \begin{equation*} M_k^{(p)}=\prod_{j=2}^k E(1+V_j)^p. \end{equation*} In particular, \begin{equation*} M_k^{(2)}=\prod_{j=2}^k (1+a_k^2)=\prod_{j=2}^k \Big(1+\frac1{k\ln k}\Big) =\exp\Big\{(1+o(1))\int_2^k\frac{dx}{x\ln x}\Big\} =(\ln k)^{1+o(1)} \end{equation*} (as $k\to\infty$). Similarly, \begin{equation*} M_k^{(4)}=\prod_{j=2}^k (1+6a_k^2+a_k^4)=(\ln k)^{6+o(1)}, \end{equation*} \begin{equation*} M_k^{(6)}=\prod_{j=2}^k (1+15a_k^2+15a_k^4+a_k^6)=(\ln k)^{15+o(1)}. \end{equation*} Hence, \begin{equation*} A^{(6)}_n=1+\sum_{k=2}^n M_{k-1}^{(6)}a_k^6\ll1+\sum_{k=2}^n (\ln k)^{15+o(1)}\frac1{k^3\ln^3k}\ll1. \end{equation*} Also, we may take \begin{equation*} B^2=\sum_{i=1}^n Var\,X_i=ES_n^2=M_n^{(2)}=(\ln n)^{1+o(1)}. \end{equation*} So, for $n\to\infty$ the right-hand side of \eqref{2} is \begin{equation*} B^4+A^{(6)}_n/B^2=(\ln n)^{2+o(1)}+O(1)/(\ln n)^{1+o(1)}=(\ln n)^{2+o(1)}, \end{equation*} whereas the left-hand side of \eqref{2} is \begin{equation*} ES_n^4=M_n^{(4)}=(\ln n)^{6+o(1)}. \end{equation*} Thus, \eqref{2} fails to hold for large enough $n$.

It remains to give

Proof of Lemma 1: Suppose the highlighted statement is true. Take any conditionally symmetric martingale difference sequence $(X_1,\dots,X_n)$ such that $\sum_{i=1}^n Var\,X_i\le B^2$. Take any real $y>0$. Let $X_{i,y}:=X_i\,1(|X_i|\le y)$ for all $i$. Then $(X_{1,y},\dots,X_{n,y})$ is a martingale difference sequence with $|X_{i,y}|\le y$ and $Var\,X_{i,y}\le Var\,X_i$ for all $i$. So, \begin{align*} P(|S_n|\ge x)&\le\sum_{i=1}^n P(|X_i|>y)+P\Big(\Big|\sum_{i=1}^nX_{i,y}\Big|\ge x\Big) \\ &\le \sum_{i=1}^n P(|X_i|>y)+2\exp\frac{-cx^2}{B^2+xy} \end{align*} by the highlighted statement, for all real $x>0$. Using this inequality with $y=B(x/B)^{2/3}$, integrating in $x>0$, and using the substitutions $z=B(x/B)^{2/3}$ and $x/B=t$, we have \begin{align*} ES_n^4&=\int_0^\infty dx\,4x^3P(|S_n|\ge x) \\ &\le\sum_{i=1}^n \int_0^\infty dx\,4x^3 P(|X_i|>B(x/B)^{2/3}) \\ & +\int_0^\infty dx\,4x^3 2\exp\frac{-cx^2}{B^2+xB(x/B)^{2/3}} \\ &\ll A^{(6)}_n/B^2+B^4. \end{align*} This completes the proof of Lemma 1 and thus the proof of Theorem 2. $\Box$

Whereas, as has just been shown, the highlighted statement is false even for conditionally symmetric martingale difference sequences $(X_1,\dots,X_n)$, note Theorem 3.6, which implies that for any conditionally symmetric martingale difference sequences $(X_1,\dots,X_n)$ such that $\sum_{i=1}^n X_i^2\le B^2$ for some real $B>0$, we have \begin{equation*} P\Big(\Big|\sum_{i=1}^n X_i\Big|\ge x\Big)\le2\exp\frac{-x^2}{2B^2} \end{equation*} for all real $x>0$.

Theorem 1: In the known exponential bounds for martingales, the conditional variances cannot be replaced by the unconditional ones.

Proof: Otherwise, we would most likely have such bounds. $\Box$ :-)

This "proof" of "Theorem 1" is not so non-serious as it may look.

Perhaps more seriously, we have

Theorem 2: The following statement is false:

There is a real constant $c>0$ such that for all natural $n$, all real $y>0$, all real $B>0$, and all martingale difference sequences $(X_1,\dots,X_n)$ such that \begin{equation*} X_i\le y \ \forall i\quad\text{and}\quad\sum_{i=1}^n Var\,X_i\le B^2\label{0}\tag{0} \end{equation*} we have \begin{equation*} P\Big(\sum_{i=1}^n X_i\ge x\Big)\le\exp\frac{-cx^2}{B^2+xy}\label{1}\tag{1} \end{equation*} for all real $x>0$.

Proof: This proof would be a bit simpler if, instead of using Corollary 2 in the Pinelis--Utev paper, you used the better bound in Theorem 3 in that paper. Indeed, one can show that, at least in the case when the $X_i$'s are conditionally symmetric (given $\mathcal F_{i-1}$), that theorem implies the Rosenthal-type inequality \begin{equation*} ES_n^4\ll B^4+A^{(4)}_n, \end{equation*} where $$S_n:=\sum_{i=1}^n X_i,$$ $a\ll b$ means $a\le Cb$ for some real $C$ depending only on $c$, and \begin{equation*} A^{(p)}_n:=\sum_{i=1}^n E|X_i|^p. \end{equation*}

Because the bound in \eqref{1} is suboptimal, it only implies an ugly version of the Rosenthal-type inequality:

Lemma 1: If the highlighted statement is true, then for conditionally symmetric martingale difference sequences $(X_1,\dots,X_n)$ such that $\sum_{i=1}^n Var\,X_i\le B^2$ we have \begin{equation*} ES_n^4\ll B^4+A^{(6)}_n/B^2. \label{2}\tag{2} \end{equation*}

This lemma will be proved at the end of this answer.

Now consider the following construction of a conditionally symmetric martingale difference sequence $(X_1,\dots,X_n)$: Let $V_1:=R_1$, where $R_1$ is a Rademacher random variable, so that $P(R_1=\pm1)=1/2$. For natural $k\ge2$, let \begin{equation*} V_k:=a_k R_k,\quad a_k:=\frac1{\sqrt{k\ln k}}, \end{equation*} where $R_2,R_3,\dots$ are independent copies of $R_1$. Let then $X_1:=V_1$, and for natural $k\ge2$ let \begin{equation*} X_k:=S_{k-1}V_k, \end{equation*} where $S_j:=\sum_{i=1}^j X_i$, as before. So, for natural $k\ge2$, \begin{equation*} S_k=S_{k-1}(1+V_k). \end{equation*} So, for any even natural $p$ and any natural $k\ge2$, we have $M_k^{(p)}:=ES_k^p=M_{k-1}^{(p)} E(1+V_k)^p$ and hence \begin{equation*} M_k^{(p)}=\prod_{j=2}^k E(1+V_j)^p. \end{equation*} In particular, \begin{equation*} M_k^{(2)}=\prod_{j=2}^k (1+a_k^2)=\prod_{j=2}^k \Big(1+\frac1{k\ln k}\Big) =\exp\Big\{(1+o(1))\int_2^k\frac{dx}{x\ln x}\Big\} =(\ln k)^{1+o(1)} \end{equation*} (as $k\to\infty$). Similarly, \begin{equation*} M_k^{(4)}=\prod_{j=2}^k (1+6a_k^2+a_k^4)=(\ln k)^{6+o(1)}, \end{equation*} \begin{equation*} M_k^{(6)}=\prod_{j=2}^k (1+15a_k^2+15a_k^4+a_k^6)=(\ln k)^{15+o(1)}. \end{equation*} Hence, \begin{equation*} A^{(6)}_n=1+\sum_{k=2}^n M_{k-1}^{(6)}a_k^6\ll1+\sum_{k=2}^n (\ln k)^{15+o(1)}\frac1{k^3\ln^3k}\ll1. \end{equation*} Also, we may take \begin{equation*} B^2=\sum_{i=1}^n Var\,X_i=ES_n^2=M_n^{(2)}=(\ln n)^{1+o(1)}. \end{equation*} So, for $n\to\infty$ the right-hand side of \eqref{2} is \begin{equation*} B^4+A^{(6)}_n/B^2=(\ln n)^{2+o(1)}+O(1)/(\ln n)^{1+o(1)}=(\ln n)^{2+o(1)}, \end{equation*} whereas the left-hand side of \eqref{2} is \begin{equation*} ES_n^4=M_n^{(4)}=(\ln n)^{6+o(1)}. \end{equation*} Thus, \eqref{2} fails to hold for large enough $n$.

It remains to give

Proof of Lemma 1: Suppose the highlighted statement is true. Take any conditionally symmetric martingale difference sequence $(X_1,\dots,X_n)$ such that $\sum_{i=1}^n Var\,X_i\le B^2$. Take any real $y>0$. Let $X_{i,y}:=X_i\,1(|X_i|\le y)$ for all $i$. Then $(X_{1,y},\dots,X_{n,y})$ is a martingale difference sequence with $|X_{i,y}|\le y$ and $Var\,X_{i,y}\le Var\,X_i$ for all $i$. So, \begin{align*} P(|S_n|\ge x)&\le\sum_{i=1}^n P(|X_i|>y)+P\Big(\Big|\sum_{i=1}^nX_{i,y}\Big|\ge x\Big) \\ &\le \sum_{i=1}^n P(|X_i|>y)+2\exp\frac{-cx^2}{B^2+xy} \end{align*} by the highlighted statement, for all real $x>0$. Using this inequality with $y=B(x/B)^{2/3}$, integrating in $x>0$, and using the substitutions $z=B(x/B)^{2/3}$ and $x/B=t$, we have \begin{align*} ES_n^4&=\int_0^\infty dx\,4x^3P(|S_n|\ge x) \\ &\le\sum_{i=1}^n \int_0^\infty dx\,4x^3 P(|X_i|>B(x/B)^{2/3}) \\ & +\int_0^\infty dx\,4x^3 2\exp\frac{-cx^2}{B^2+xB(x/B)^{2/3}} \\ &\ll A^{(6)}_n/B^2+B^4. \end{align*} This completes the proof of Lemma 1 and thus the proof of Theorem 2. $\Box$

Whereas, as has just been shown, the highlighted statement is false even for conditionally symmetric martingale difference sequences $(X_1,\dots,X_n)$, note Theorem 3.6, which implies that for any conditionally symmetric martingale difference sequences $(X_1,\dots,X_n)$ such that $\sum_{i=1}^n X_i^2\le B^2$ for some real $B>0$, we have \begin{equation*} P\Big(\Big|\sum_{i=1}^n X_i\Big|\ge x\Big)\le2\exp\frac{-x^2}{2B^2} \end{equation*} for all real $x>0$.

Theorem 1: In the known exponential bounds for martingales, the conditional variances cannot be replaced by the unconditional ones.

Proof: Otherwise, we would most likely have such bounds. $\Box$ :-)

This "proof" of "Theorem 1" is not so non-serious as it may look.

Perhaps more seriously, we have

Theorem 2: The following statement is false:

There is a real constant $c>0$ such that for all natural $n$, all real $y>0$, all real $B>0$, and all martingale difference sequences $(X_1,\dots,X_n)$ such that \begin{equation*} X_i\le y \ \forall i\quad\text{and}\quad\sum_{i=1}^n Var\,X_i\le B^2\label{0}\tag{0} \end{equation*} we have \begin{equation*} P\Big(\sum_{i=1}^n X_i\ge x\Big)\le\exp\frac{-cx^2}{B^2+xy}\label{1}\tag{1} \end{equation*} for all real $x>0$.

Proof: This proof would be a bit simpler if, instead of using Corollary 2 in the Pinelis--Utev paper, you used the better bound in Theorem 3 in that paper. Indeed, one can show that that theorem implies the Rosenthal-type inequality \begin{equation*} ES_n^4\ll B^4+A^{(4)}_n, \end{equation*} where $$S_n:=\sum_{i=1}^n X_i,$$ $a\ll b$ means $a\le Cb$ for some real $C$ depending only on $c$, and \begin{equation*} A^{(p)}_n:=\sum_{i=1}^n E|X_i|^p. \end{equation*}

Because the bound in \eqref{1} is suboptimal, it only implies an ugly version of the Rosenthal-type inequality:

Lemma 1: If the highlighted statement is true, then for martingale difference sequences $(X_1,\dots,X_n)$ such that $\sum_{i=1}^n Var\,X_i\le B^2$ we have \begin{equation*} ES_n^4\ll B^4+A^{(6)}_n/B^2. \label{2}\tag{2} \end{equation*}

This lemma will be proved at the end of this answer.

Now consider the following construction: Let $V_1:=R_1$, where $R_1$ is a Rademacher random variable, so that $P(R_1=\pm1)=1/2$. For natural $k\ge2$, let \begin{equation*} V_k:=a_k R_k,\quad a_k:=\frac1{\sqrt{k\ln k}}, \end{equation*} where $R_2,R_3,\dots$ are independent copies of $R_1$. Let then $X_1:=V_1$ and for natural $k\ge2$, let \begin{equation*} X_k:=S_{k-1}V_k, \end{equation*} where $S_j:=\sum_{i=1}^j X_i$, as before. So, for natural $k\ge2$, \begin{equation*} S_k=S_{k-1}(1+V_k). \end{equation*} So, for any even natural $p$ and any natural $k\ge2$, we have $M_k^{(p)}:=ES_k^p=M_{k-1}^{(p)} E(1+V_k)^p$ and hence \begin{equation*} M_k^{(p)}=\prod_{j=2}^k E(1+V_j)^p. \end{equation*} In particular, \begin{equation*} M_k^{(2)}=\prod_{j=2}^k (1+a_k^2)=\prod_{j=2}^k \Big(1+\frac1{k\ln k}\Big) =\exp\Big\{(1+o(1))\int_2^k\frac{dx}{x\ln x}\Big\} =(\ln k)^{1+o(1)} \end{equation*} (as $k\to\infty$). Similarly, \begin{equation*} M_k^{(4)}=\prod_{j=2}^k (1+6a_k^2+a_k^4)=(\ln k)^{6+o(1)}, \end{equation*} \begin{equation*} M_k^{(6)}=\prod_{j=2}^k (1+15a_k^2+15a_k^4+a_k^6)=(\ln k)^{15+o(1)}. \end{equation*} Hence, \begin{equation*} A^{(6)}_n=1+\sum_{k=2}^n M_{k-1}^{(6)}a_k^6\ll1+\sum_{k=2}^n (\ln k)^{15+o(1)}\frac1{k^3\ln^3k}\ll1. \end{equation*} Also, we may take \begin{equation*} B^2=\sum_{i=1}^n Var\,X_i=ES_n^2=M_n^{(2)}=(\ln n)^{1+o(1)}. \end{equation*} So, for $n\to\infty$ the right-hand side of \eqref{2} is \begin{equation*} B^4+A^{(6)}_n/B^2=(\ln n)^{2+o(1)}+O(1)/(\ln n)^{1+o(1)}=(\ln n)^{2+o(1)}, \end{equation*} whereas the left-hand side of \eqref{2} is \begin{equation*} ES_n^4=M_n^{(4)}=(\ln n)^{6+o(1)}. \end{equation*} Thus, \eqref{2} fails to hold for large enough $n$.

It remains to give

Proof of Lemma 1: Suppose the highlighted statement is true. Take any martingale difference sequence $(X_1,\dots,X_n)$ such that $\sum_{i=1}^n Var\,X_i\le B^2$. Take any real $y>0$. Let $X_{i,y}:=\min(y,X_i)$ for all $i$. Then $(X_{1,y},\dots,X_{n,y})$ is a supermartingale difference sequence and $Var\,X_{i,y}\le Var\,X_i$. So, \begin{align*} P(S_n\ge x)&\le\sum_{i=1}^n P(X_i>y)+P\Big(\sum_{i=1}^nX_{i,y}\ge x\Big) \\ &\le\sum_{i=1}^n P(X_i>y)+P\Big(\sum_{i=1}^n[X_{i,y}-E(X_{i,y}|\mathcal F_{i-1})]\ge x\Big) \\ &\le \sum_{i=1}^n P(X_i>y)+\exp\frac{-cx^2}{B^2+xy} \end{align*} by the highlighted statement, for all real $x>0$. So, \begin{equation*} P(|S_n|\ge x)\le\sum_{i=1}^n P(|X_i|>y)+2\exp\frac{-cx^2}{B^2+xy}. \end{equation*} Using this inequality with $y=B(x/B)^{2/3}$, integrating in $x>0$, and using the substitutions $z=B(x/B)^{2/3}$ and $x/B=t$, we have \begin{align*} ES_n^4&=\int_0^\infty dx\,4x^3P(|S_n|\ge x) \\ &\le\sum_{i=1}^n \int_0^\infty dx\,4x^3 P(|X_i|>B(x/B)^{2/3}) \\ & +\int_0^\infty dx\,4x^3 2\exp\frac{-cx^2}{B^2+xB(x/B)^{2/3}} \\ &\ll A^{(6)}_n/B^2+B^4. \end{align*} This completes the proof of Lemma 1 and thus the proof of Theorem 2. $\Box$

Theorem 1: In the known exponential bounds for martingales, the conditional variances cannot be replaced by the unconditional ones.

Proof: Otherwise, we would most likely have such bounds. $\Box$ :-)

This "proof" of "Theorem 1" is not so non-serious as it may look.

Perhaps more seriously, we have

Theorem 2: The following statement is false:

There is a real constant $c>0$ such that for all natural $n$, all real $y>0$, all real $B>0$, and all martingale difference sequences $(X_1,\dots,X_n)$ such that \begin{equation*} X_i\le y \ \forall i\quad\text{and}\quad\sum_{i=1}^n Var\,X_i\le B^2\label{0}\tag{0} \end{equation*} we have \begin{equation*} P\Big(\sum_{i=1}^n X_i\ge x\Big)\le\exp\frac{-cx^2}{B^2+xy}\label{1}\tag{1} \end{equation*} for all real $x>0$.

Proof: This proof would be a bit simpler if, instead of using Corollary 2 in the Pinelis--Utev paper, you used the better bound in Theorem 3 in that paper. Indeed, one can show that, at least in the case when the $X_i$'s are conditionally symmetric (given $\mathcal F_{i-1}$), that theorem implies the Rosenthal-type inequality \begin{equation*} ES_n^4\ll B^4+A^{(4)}_n, \end{equation*} where $$S_n:=\sum_{i=1}^n X_i,$$ $a\ll b$ means $a\le Cb$ for some real $C$ depending only on $c$, and \begin{equation*} A^{(p)}_n:=\sum_{i=1}^n E|X_i|^p. \end{equation*}

Because the bound in \eqref{1} is suboptimal, it only implies an ugly version of the Rosenthal-type inequality:

Lemma 1: If the highlighted statement is true, then for conditionally symmetric martingale difference sequences $(X_1,\dots,X_n)$ such that $\sum_{i=1}^n Var\,X_i\le B^2$ we have \begin{equation*} ES_n^4\ll B^4+A^{(6)}_n/B^2. \label{2}\tag{2} \end{equation*}

This lemma will be proved at the end of this answer.

Now consider the following construction of a conditionally symmetric martingale difference sequence $(X_1,\dots,X_n)$: Let $V_1:=R_1$, where $R_1$ is a Rademacher random variable, so that $P(R_1=\pm1)=1/2$. For natural $k\ge2$, let \begin{equation*} V_k:=a_k R_k,\quad a_k:=\frac1{\sqrt{k\ln k}}, \end{equation*} where $R_2,R_3,\dots$ are independent copies of $R_1$. Let then $X_1:=V_1$, and for natural $k\ge2$ let \begin{equation*} X_k:=S_{k-1}V_k, \end{equation*} where $S_j:=\sum_{i=1}^j X_i$, as before. So, for natural $k\ge2$, \begin{equation*} S_k=S_{k-1}(1+V_k). \end{equation*} So, for any even natural $p$ and any natural $k\ge2$, we have $M_k^{(p)}:=ES_k^p=M_{k-1}^{(p)} E(1+V_k)^p$ and hence \begin{equation*} M_k^{(p)}=\prod_{j=2}^k E(1+V_j)^p. \end{equation*} In particular, \begin{equation*} M_k^{(2)}=\prod_{j=2}^k (1+a_k^2)=\prod_{j=2}^k \Big(1+\frac1{k\ln k}\Big) =\exp\Big\{(1+o(1))\int_2^k\frac{dx}{x\ln x}\Big\} =(\ln k)^{1+o(1)} \end{equation*} (as $k\to\infty$). Similarly, \begin{equation*} M_k^{(4)}=\prod_{j=2}^k (1+6a_k^2+a_k^4)=(\ln k)^{6+o(1)}, \end{equation*} \begin{equation*} M_k^{(6)}=\prod_{j=2}^k (1+15a_k^2+15a_k^4+a_k^6)=(\ln k)^{15+o(1)}. \end{equation*} Hence, \begin{equation*} A^{(6)}_n=1+\sum_{k=2}^n M_{k-1}^{(6)}a_k^6\ll1+\sum_{k=2}^n (\ln k)^{15+o(1)}\frac1{k^3\ln^3k}\ll1. \end{equation*} Also, we may take \begin{equation*} B^2=\sum_{i=1}^n Var\,X_i=ES_n^2=M_n^{(2)}=(\ln n)^{1+o(1)}. \end{equation*} So, for $n\to\infty$ the right-hand side of \eqref{2} is \begin{equation*} B^4+A^{(6)}_n/B^2=(\ln n)^{2+o(1)}+O(1)/(\ln n)^{1+o(1)}=(\ln n)^{2+o(1)}, \end{equation*} whereas the left-hand side of \eqref{2} is \begin{equation*} ES_n^4=M_n^{(4)}=(\ln n)^{6+o(1)}. \end{equation*} Thus, \eqref{2} fails to hold for large enough $n$.

It remains to give

Proof of Lemma 1: Suppose the highlighted statement is true. Take any conditionally symmetric martingale difference sequence $(X_1,\dots,X_n)$ such that $\sum_{i=1}^n Var\,X_i\le B^2$. Take any real $y>0$. Let $X_{i,y}:=X_i\,1(|X_i|\le y)$ for all $i$. Then $(X_{1,y},\dots,X_{n,y})$ is a martingale difference sequence with $|X_{i,y}|\le y$ and $Var\,X_{i,y}\le Var\,X_i$ for all $i$. So, \begin{align*} P(|S_n|\ge x)&\le\sum_{i=1}^n P(|X_i|>y)+P\Big(\Big|\sum_{i=1}^nX_{i,y}\Big|\ge x\Big) \\ &\le \sum_{i=1}^n P(|X_i|>y)+2\exp\frac{-cx^2}{B^2+xy} \end{align*} by the highlighted statement, for all real $x>0$. Using this inequality with $y=B(x/B)^{2/3}$, integrating in $x>0$, and using the substitutions $z=B(x/B)^{2/3}$ and $x/B=t$, we have \begin{align*} ES_n^4&=\int_0^\infty dx\,4x^3P(|S_n|\ge x) \\ &\le\sum_{i=1}^n \int_0^\infty dx\,4x^3 P(|X_i|>B(x/B)^{2/3}) \\ & +\int_0^\infty dx\,4x^3 2\exp\frac{-cx^2}{B^2+xB(x/B)^{2/3}} \\ &\ll A^{(6)}_n/B^2+B^4. \end{align*} This completes the proof of Lemma 1 and thus the proof of Theorem 2. $\Box$

Whereas, as has just been shown, the highlighted statement is false even for conditionally symmetric martingale difference sequences $(X_1,\dots,X_n)$, note Theorem 3.6, which implies that for any conditionally symmetric martingale difference sequences $(X_1,\dots,X_n)$ such that $\sum_{i=1}^n X_i^2\le B^2$ for some real $B>0$, we have \begin{equation*} P\Big(\Big|\sum_{i=1}^n X_i\Big|\ge x\Big)\le2\exp\frac{-x^2}{2B^2} \end{equation*} for all real $x>0$.

Theorem 1: In the known exponential bounds for martingales, the conditional variances cannot be replaced by the unconditional ones.

Proof: Otherwise, we would most likely have such bounds. $\Box$ :-)

This "proof" of "Theorem 1" is not so non-serious as it may look.

Perhaps more seriously, we have

Theorem 2: The following statement is false:

There is a real constant $c>0$ such that for all natural $n$, all real $y>0$, all real $B>0$, and all martingale difference sequences $(X_1,\dots,X_n)$ such that \begin{equation*} X_i\le y \ \forall i\quad\text{and}\quad\sum_{i=1}^n Var\,X_i\le B^2\label{0}\tag{0} \end{equation*} we have \begin{equation*} P\Big(\sum_{i=1}^n X_i\ge x\Big)\le\exp\frac{-cx^2}{B^2+xy}\label{1}\tag{1} \end{equation*} for all real $x>0$.

Proof: This proof would be a bit simpler if, instead of using Corollary 2 in the Pinelis--Utev paper, you used the better bound in Theorem 3 in that paper. Indeed, one can show that that theorem implies the Rosenthal-type inequality \begin{equation*} ES_n^4\ll B^4+A^{(4)}_n, \end{equation*} where $$S_n:=\sum_{i=1}^n X_i,$$ $a\ll b$ means $a\le Cb$ for some real $C$ depending only on $c$, and \begin{equation*} A^{(p)}_n:=\sum_{i=1}^n E|X_i|^p. \end{equation*}

Because the bound in \eqref{1} is suboptimal, it only implies an ugly version of the Rosenthal-type inequality:

Lemma 1: If the highlighted statement is true, then for martingale difference sequences $(X_1,\dots,X_n)$ such that $\sum_{i=1}^n Var\,X_i\le B^2$ we have \begin{equation*} ES_n^4\ll B^4+A^{(6)}_n/B^2. \label{2}\tag{2} \end{equation*}

This lemma will be proved at the end of this answer.

Now consider the following construction: Let $V_1:=R_1$, where $R_1$ is a Rademacher random variable, so that $P(R_1=\pm1)=1/2$. For natural $k\ge2$, let \begin{equation*} V_k:=a_k R_k,\quad a_k:=\frac1{\sqrt{k\ln k}}, \end{equation*} where $R_2,R_3,\dots$ are independent copies of $R_1$. Let then $X_1:=V_1$ and for natural $k\ge2$, let \begin{equation*} X_k:=S_{k-1}V_k, \end{equation*} where $S_j:=\sum_{i=1}^j X_i$, as before. So, for natural $k\ge2$, \begin{equation*} S_k=S_{k-1}(1+V_k). \end{equation*} So, for any even natural $p$ and any natural $k\ge2$, we have $M_k^{(p)}:=ES_k^p=M_{k-1}^{(p)} E(1+V_k)^p$ and hence \begin{equation*} M_k^{(p)}=\prod_{j=1}^k E(1+V_j)^p. \end{equation*} In particular, \begin{equation*} M_k^{(2)}=\prod_{j=1}^k (1+a_k^2)=\prod_{j=1}^k \Big(1+\frac1{k\ln k}\Big) =\exp\Big\{(1+o(1))\int_2^k\frac{dx}{x\ln x}\Big\} =(\ln k)^{1+o(1)} \end{equation*} (as $k\to\infty$). Similarly, \begin{equation*} M_k^{(4)}=\prod_{j=1}^k (1+6a_k^2+a_k^4)=(\ln k)^{6+o(1)}, \end{equation*} \begin{equation*} M_k^{(6)}=\prod_{j=1}^k (1+15a_k^2+15a_k^4+a_k^6)=(\ln k)^{15+o(1)}. \end{equation*} Hence, \begin{equation*} A^{(6)}_n=\sum_{k=1}^n M_{k-1}^{(6)}a_k^6\ll\sum_{k=1}^n (\ln k)^{15+o(1)}\frac1{k^3\ln^3k}\ll1. \end{equation*} Also, we may take \begin{equation*} B^2=\sum_{i=1}^n Var\,X_i=ES_n^2=M_n^{(2)}=(\ln n)^{1+o(1)}. \end{equation*} So, for $n\to\infty$ the right-hand side of \eqref{2} is \begin{equation*} B^4+A^{(6)}_n/B^2=(\ln n)^{2+o(1)}+O(1)/(\ln n)^{1+o(1)}=(\ln n)^{2+o(1)}, \end{equation*} whereas the left-hand side of \eqref{2} is \begin{equation*} ES_n^4=M_k^{(4)}=(\ln k)^{6+o(1)}. \end{equation*} Thus, \eqref{2} fails to hold.

It remains to give

Proof of Lemma 1: Suppose the highlighted statement is true. Take any martingale difference sequence $(X_1,\dots,X_n)$ such that $\sum_{i=1}^n Var\,X_i\le B^2$. Take any real $y>0$. Let $X_{i,y}:=\min(y,X_i)$ for all $i$. Then $(X_{1,y},\dots,X_{n,y})$ is a supermartingale difference sequence and $Var\,X_{i,y}\le Var\,X_i$. So, \begin{align*} P(S_n\ge x)&\le\sum_{i=1}^n P(X_i>y)+P\Big(\sum_{i=1}^nX_{i,y}\ge x\Big) \\ &\le\sum_{i=1}^n P(X_i>y)+P\Big(\sum_{i=1}^n[X_{i,y}-E(X_{i,y}|\mathcal F_{i-1})]\ge x\Big) \\ &\le \sum_{i=1}^n P(X_i>y)+\exp\frac{-cx^2}{B^2+xy} \end{align*} by the highlighted statement, for all real $x>0$. So, \begin{equation*} P(|S_n|\ge x)\le\sum_{i=1}^n P(|X_i|>y)+2\exp\frac{-cx^2}{B^2+xy}. \end{equation*} Using this inequality with $y=B(x/B)^{2/3}$, integrating in $x>0$, and using the substitutions $z=B(x/B)^{2/3}$ and $x/B=t$, we have \begin{align*} ES_n^4&=\int_0^\infty dx\,4x^3P(|S_n|\ge x) \\ &\le\sum_{i=1}^n \int_0^\infty dx\,4x^3 P(|X_i|>B(x/B)^{2/3}) \\ & +\int_0^\infty dx\,4x^3 2\exp\frac{-cx^2}{B^2+xB(x/B)^{2/3}} \\ &\ll A^{(6)}_n/B^2+B^4. \end{align*} This completes the proof of Lemma 1 and thus the proof of Theorem 2. $\Box$

Theorem 1: In the known exponential bounds for martingales, the conditional variances cannot be replaced by the unconditional ones.

Proof: Otherwise, we would most likely have such bounds. $\Box$ :-)

This "proof" of "Theorem 1" is not so non-serious as it may look.

Perhaps more seriously, we have

Theorem 2: The following statement is false:

There is a real constant $c>0$ such that for all natural $n$, all real $y>0$, all real $B>0$, and all martingale difference sequences $(X_1,\dots,X_n)$ such that \begin{equation*} X_i\le y \ \forall i\quad\text{and}\quad\sum_{i=1}^n Var\,X_i\le B^2\label{0}\tag{0} \end{equation*} we have \begin{equation*} P\Big(\sum_{i=1}^n X_i\ge x\Big)\le\exp\frac{-cx^2}{B^2+xy}\label{1}\tag{1} \end{equation*} for all real $x>0$.

Proof: This proof would be a bit simpler if, instead of using Corollary 2 in the Pinelis--Utev paper, you used the better bound in Theorem 3 in that paper. Indeed, one can show that that theorem implies the Rosenthal-type inequality \begin{equation*} ES_n^4\ll B^4+A^{(4)}_n, \end{equation*} where $$S_n:=\sum_{i=1}^n X_i,$$ $a\ll b$ means $a\le Cb$ for some real $C$ depending only on $c$, and \begin{equation*} A^{(p)}_n:=\sum_{i=1}^n E|X_i|^p. \end{equation*}

Because the bound in \eqref{1} is suboptimal, it only implies an ugly version of the Rosenthal-type inequality:

Lemma 1: If the highlighted statement is true, then for martingale difference sequences $(X_1,\dots,X_n)$ such that $\sum_{i=1}^n Var\,X_i\le B^2$ we have \begin{equation*} ES_n^4\ll B^4+A^{(6)}_n/B^2. \label{2}\tag{2} \end{equation*}

This lemma will be proved at the end of this answer.

Now consider the following construction: Let $V_1:=R_1$, where $R_1$ is a Rademacher random variable, so that $P(R_1=\pm1)=1/2$. For natural $k\ge2$, let \begin{equation*} V_k:=a_k R_k,\quad a_k:=\frac1{\sqrt{k\ln k}}, \end{equation*} where $R_2,R_3,\dots$ are independent copies of $R_1$. Let then $X_1:=V_1$ and for natural $k\ge2$, let \begin{equation*} X_k:=S_{k-1}V_k, \end{equation*} where $S_j:=\sum_{i=1}^j X_i$, as before. So, for natural $k\ge2$, \begin{equation*} S_k=S_{k-1}(1+V_k). \end{equation*} So, for any even natural $p$ and any natural $k\ge2$, we have $M_k^{(p)}:=ES_k^p=M_{k-1}^{(p)} E(1+V_k)^p$ and hence \begin{equation*} M_k^{(p)}=\prod_{j=2}^k E(1+V_j)^p. \end{equation*} In particular, \begin{equation*} M_k^{(2)}=\prod_{j=2}^k (1+a_k^2)=\prod_{j=2}^k \Big(1+\frac1{k\ln k}\Big) =\exp\Big\{(1+o(1))\int_2^k\frac{dx}{x\ln x}\Big\} =(\ln k)^{1+o(1)} \end{equation*} (as $k\to\infty$). Similarly, \begin{equation*} M_k^{(4)}=\prod_{j=2}^k (1+6a_k^2+a_k^4)=(\ln k)^{6+o(1)}, \end{equation*} \begin{equation*} M_k^{(6)}=\prod_{j=2}^k (1+15a_k^2+15a_k^4+a_k^6)=(\ln k)^{15+o(1)}. \end{equation*} Hence, \begin{equation*} A^{(6)}_n=1+\sum_{k=2}^n M_{k-1}^{(6)}a_k^6\ll1+\sum_{k=2}^n (\ln k)^{15+o(1)}\frac1{k^3\ln^3k}\ll1. \end{equation*} Also, we may take \begin{equation*} B^2=\sum_{i=1}^n Var\,X_i=ES_n^2=M_n^{(2)}=(\ln n)^{1+o(1)}. \end{equation*} So, for $n\to\infty$ the right-hand side of \eqref{2} is \begin{equation*} B^4+A^{(6)}_n/B^2=(\ln n)^{2+o(1)}+O(1)/(\ln n)^{1+o(1)}=(\ln n)^{2+o(1)}, \end{equation*} whereas the left-hand side of \eqref{2} is \begin{equation*} ES_n^4=M_n^{(4)}=(\ln n)^{6+o(1)}. \end{equation*} Thus, \eqref{2} fails to hold for large enough $n$.

It remains to give

Proof of Lemma 1: Suppose the highlighted statement is true. Take any martingale difference sequence $(X_1,\dots,X_n)$ such that $\sum_{i=1}^n Var\,X_i\le B^2$. Take any real $y>0$. Let $X_{i,y}:=\min(y,X_i)$ for all $i$. Then $(X_{1,y},\dots,X_{n,y})$ is a supermartingale difference sequence and $Var\,X_{i,y}\le Var\,X_i$. So, \begin{align*} P(S_n\ge x)&\le\sum_{i=1}^n P(X_i>y)+P\Big(\sum_{i=1}^nX_{i,y}\ge x\Big) \\ &\le\sum_{i=1}^n P(X_i>y)+P\Big(\sum_{i=1}^n[X_{i,y}-E(X_{i,y}|\mathcal F_{i-1})]\ge x\Big) \\ &\le \sum_{i=1}^n P(X_i>y)+\exp\frac{-cx^2}{B^2+xy} \end{align*} by the highlighted statement, for all real $x>0$. So, \begin{equation*} P(|S_n|\ge x)\le\sum_{i=1}^n P(|X_i|>y)+2\exp\frac{-cx^2}{B^2+xy}. \end{equation*} Using this inequality with $y=B(x/B)^{2/3}$, integrating in $x>0$, and using the substitutions $z=B(x/B)^{2/3}$ and $x/B=t$, we have \begin{align*} ES_n^4&=\int_0^\infty dx\,4x^3P(|S_n|\ge x) \\ &\le\sum_{i=1}^n \int_0^\infty dx\,4x^3 P(|X_i|>B(x/B)^{2/3}) \\ & +\int_0^\infty dx\,4x^3 2\exp\frac{-cx^2}{B^2+xB(x/B)^{2/3}} \\ &\ll A^{(6)}_n/B^2+B^4. \end{align*} This completes the proof of Lemma 1 and thus the proof of Theorem 2. $\Box$