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Welcome to MathOverflow! The answer to your question is yes. Indeed, let $X$ and $Y$ be independent random elements of the set $\{0,1\}^n$ whose distributions are percolation measures, which latter thus satisfy the Harris--FKG inequality, Proposition 2.3: $Ef(X)g(X)\ge Ef(X)\, Eg(X)$ and $Ef(Y)g(Y)\ge Ef(Y)\, Eg(Y)$ for all bounded nondecreasing functions $f$ and $g$. We have to show that for any such functions $f$ and $g$ we have $Ef(X\vee Y)g(X\vee Y)\ge Ef(X\vee Y)\, Eg(X\vee Y)$, where $X\vee Y$ is coordinate-wise maximum of $X$ and $Y$.

Letting $F(y):=Ef(X\vee y)$ and $G(y):=Eg(X\vee y)$ for $y\in\{0,1\}^n$, we have \begin{align} Ef(X\vee Y)g(X\vee Y)&=\sum_{y\in\{0,1\}^n} P(Y=y)Ef(X\vee y)g(X\vee y) \\ &\ge\sum_{y\in\{0,1\}^n} P(Y=y)Ef(X\vee y)\,Eg(X\vee y) \\ &=\sum_{y\in\{0,1\}^n} P(Y=y)F(y)\,G(y) \\ &=EF(Y)G(Y) \\ &\ge EF(Y)\,EG(Y) \\ &= Ef(X\vee Y)\, Eg(X\vee Y), \end{align} as desired. The first inequality in the above multiline display follows by the Harris--FKG inequality for $X$ and nondecreasing functions $x\mapsto f(x\vee y)$ and $x\mapsto g(x\vee y)$, and the second inequality there is an instance of the Harris--FKG inequality for $Y$ and nondecreasing functions $F$ and $G$.

Welcome to MathOverflow! The answer to your question is yes. Indeed, let $X$ and $Y$ be independent random elements of the set $\{0,1\}^n$ whose distributions are percolation measures, which latter thus satisfy the Harris--FKG inequality, Proposition 2.3: $Ef(X)g(X)\ge Ef(X)\, Eg(X)$ and $Ef(Y)g(Y)\ge Ef(Y)\, Eg(Y)$ for all bounded nondecreasing functions $f$ and $g$. We have to show that for any such functions $f$ and $g$ we have $$Ef(X\vee Y)g(X\vee Y)\ge Ef(X\vee Y)\, Eg(X\vee Y),\tag{1}$$ where $X\vee Y$ is coordinate-wise maximum of $X$ and $Y$. Note that (i) your union of the measures is the distribution of $X\vee Y$ and (ii) the inequality $P(X\vee Y\in A\cap B)\ge P(X\vee Y\in A)P(X\vee Y\in B)$ for increasing sets $A$ and $B$ is a special case of (1), with $f$ and $g$ being the indicator functions of $A$ and $B$.

To prove (1), let $F(y):=Ef(X\vee y)$ and $G(y):=Eg(X\vee y)$ for $y\in\{0,1\}^n$. Then \begin{align} Ef(X\vee Y)g(X\vee Y)&=\sum_{y\in\{0,1\}^n} P(Y=y)Ef(X\vee y)g(X\vee y) \\ &\ge\sum_{y\in\{0,1\}^n} P(Y=y)Ef(X\vee y)\,Eg(X\vee y) \\ &=\sum_{y\in\{0,1\}^n} P(Y=y)F(y)\,G(y) \\ &=EF(Y)G(Y) \\ &\ge EF(Y)\,EG(Y) \\ &= Ef(X\vee Y)\, Eg(X\vee Y), \end{align} so that (1) is proved. The first inequality in the above multiline display follows by the Harris--FKG inequality for $X$ and nondecreasing functions $x\mapsto f(x\vee y)$ and $x\mapsto g(x\vee y)$, and the second inequality there is an instance of the Harris--FKG inequality for $Y$ and nondecreasing functions $F$ and $G$.

Welcome to MathOverflow! The answer to your question is yes. Indeed, let $X$ and $Y$ be independent random elements of the set $\{0,1\}^n$ whose distributions are percolation measures, which latter thus satisfy the Harris--FKG inequality, Proposition 2.3: $Ef(X)g(X)\ge Ef(X)\, Eg(X)$ and $Ef(Y)g(Y)\ge Ef(Y)\, Eg(Y)$ for all bounded nondecreasing functions $f$ and $g$. We have to show that for any such functions $f$ and $g$ we have $Ef(X\vee Y)g(X\vee Y)\ge Ef(X\vee Y)\, Eg(X\vee Y)$, where $X\vee Y$ is coordinate-wise maximum of $X$ and $Y$.

Letting $F(y):=Ef(X\vee y)$ and $G(y):=Eg(X\vee y)$ for $y\in\{0,1\}^n$, we have \begin{align} Ef(X\vee Y)g(X\vee Y)&=\sum_{y\in\{0,1\}^n} P(Y=y)Ef(X\vee y)g(X\vee y) \\ &\ge\sum_{y\in\{0,1\}^n} P(Y=y)Ef(X\vee y)\,Eg(X\vee y) \\ &=\sum_{y\in\{0,1\}^n} P(Y=y)F(y)\,G(y) \\ &=EF(Y)G(Y) \\ &\ge EF(Y)\,EG(Y) \\ &= Ef(X\vee Y)\, Eg(X\vee Y), \end{align} as desired. The first inequality in the above multiline display is an instance of the Harris--FKG inequality for $X$ and nondecreasing functions $x\mapsto f(x\vee y)$ and $x\mapsto g(x\vee y)$, and the second inequality there is an instance of the Harris--FKG inequality for $Y$ and nondecreasing functions $F$ and $G$.

Welcome to MathOverflow! The answer to your question is yes. Indeed, let $X$ and $Y$ be independent random elements of the set $\{0,1\}^n$ whose distributions are percolation measures, which latter thus satisfy the Harris--FKG inequality, Proposition 2.3: $Ef(X)g(X)\ge Ef(X)\, Eg(X)$ and $Ef(Y)g(Y)\ge Ef(Y)\, Eg(Y)$ for all bounded nondecreasing functions $f$ and $g$. We have to show that for any such functions $f$ and $g$ we have $Ef(X\vee Y)g(X\vee Y)\ge Ef(X\vee Y)\, Eg(X\vee Y)$, where $X\vee Y$ is coordinate-wise maximum of $X$ and $Y$.

Letting $F(y):=Ef(X\vee y)$ and $G(y):=Eg(X\vee y)$ for $y\in\{0,1\}^n$, we have \begin{align} Ef(X\vee Y)g(X\vee Y)&=\sum_{y\in\{0,1\}^n} P(Y=y)Ef(X\vee y)g(X\vee y) \\ &\ge\sum_{y\in\{0,1\}^n} P(Y=y)Ef(X\vee y)\,Eg(X\vee y) \\ &=\sum_{y\in\{0,1\}^n} P(Y=y)F(y)\,G(y) \\ &=EF(Y)G(Y) \\ &\ge EF(Y)\,EG(Y) \\ &= Ef(X\vee Y)\, Eg(X\vee Y), \end{align} as desired. The first inequality in the above multiline display follows by the Harris--FKG inequality for $X$ and nondecreasing functions $x\mapsto f(x\vee y)$ and $x\mapsto g(x\vee y)$, and the second inequality there is an instance of the Harris--FKG inequality for $Y$ and nondecreasing functions $F$ and $G$.

Welcome to MathOverflow! The answer to your question is yes. Indeed, let $X$ and $Y$ be independent random elements of the set $\{0,1\}^n$ whose distributions are percolation measures, which latter thus satisfy the Harris--FKG inequality, Proposition 2.3: $Ef(X)g(X)\ge Ef(X)\, Eg(X)$ and $Ef(Y)g(Y)\ge Ef(Y)\, Eg(Y)$ for all bounded nondecreasing functions $f$ and $g$. We have to show that for any such functions $f$ and $g$ we have $Ef(X\vee Y)g(X\vee Y)\ge Ef(X\vee Y)\, Eg(X\vee Y)$, where $X\vee Y$ is coordinate-wise maximum of $X$ and $Y$.

Letting $F(y):=Ef(X\vee y)$ and $G(y):=Eg(X\vee y)$ for $y\in\{0,1\}^n$, we have \begin{align} Ef(X\vee Y)g(X\vee Y)&=\sum_{y\in\{0,1\}^n} P(Y=y)Ef(X\vee y)g(X\vee y) \\ &\ge\sum_{y\in\{0,1\}^n} P(Y=y)Ef(X\vee y)\,Eg(X\vee y) \\ &=\sum_{y\in\{0,1\}^n} P(Y=y)F(y)\,G(y) \\ &=EF(Y)G(Y) \\ &\ge EF(Y)\,EG(Y) \\ &= Ef(X\vee Y)\, Eg(X\vee Y), \end{align} as desired. the first inequality in the above multiline display is an instance of the Harris--FKG inequality for $X$ and nondecreasing functions $x\mapsto f(x\vee y)$ and $x\mapsto g(x\vee y)$, and the second inequality there is an instance of the Harris--FKG inequality for $Y$ and nondecreasing functions $F$ and $G$.

Welcome to MathOverflow! The answer to your question is yes. Indeed, let $X$ and $Y$ be independent random elements of the set $\{0,1\}^n$ whose distributions are percolation measures, which latter thus satisfy the Harris--FKG inequality, Proposition 2.3: $Ef(X)g(X)\ge Ef(X)\, Eg(X)$ and $Ef(Y)g(Y)\ge Ef(Y)\, Eg(Y)$ for all bounded nondecreasing functions $f$ and $g$. We have to show that for any such functions $f$ and $g$ we have $Ef(X\vee Y)g(X\vee Y)\ge Ef(X\vee Y)\, Eg(X\vee Y)$, where $X\vee Y$ is coordinate-wise maximum of $X$ and $Y$.

Letting $F(y):=Ef(X\vee y)$ and $G(y):=Eg(X\vee y)$ for $y\in\{0,1\}^n$, we have \begin{align} Ef(X\vee Y)g(X\vee Y)&=\sum_{y\in\{0,1\}^n} P(Y=y)Ef(X\vee y)g(X\vee y) \\ &\ge\sum_{y\in\{0,1\}^n} P(Y=y)Ef(X\vee y)\,Eg(X\vee y) \\ &=\sum_{y\in\{0,1\}^n} P(Y=y)F(y)\,G(y) \\ &=EF(Y)G(Y) \\ &\ge EF(Y)\,EG(Y) \\ &= Ef(X\vee Y)\, Eg(X\vee Y), \end{align} as desired. The first inequality in the above multiline display is an instance of the Harris--FKG inequality for $X$ and nondecreasing functions $x\mapsto f(x\vee y)$ and $x\mapsto g(x\vee y)$, and the second inequality there is an instance of the Harris--FKG inequality for $Y$ and nondecreasing functions $F$ and $G$.