Suppose that $(X,d)$ is a separable metric space and that $\alpha,\beta>0$. If $P$ and $Q$ are probability measures on $X$ satisfying $$ P(E) \le Q(E^\alpha) + \beta $$ for any Borel-measurable set $E \subset X$, then for any $\varepsilon>0$ there exist two nonnegative measures $\nu,\gamma$ on $X\times X$ such that

1. $\mu:=\gamma + \nu$ is a law on $X\times X$ with marginals $P$ and $Q$.

2. $\nu \{(x,y)\in X\times X: d(x,y) > \alpha + \varepsilon \} = 0$

3. $\gamma(X\times X) \le \beta + \varepsilon$

I want to prove that if real random variables X,Y on $(\Omega,\mathbb{P})$ have P,Q as distributions then

$$\mathbb{P}(d(X,Y) <\alpha )\geq \mu(d(x,y) < \alpha ).$$

Can I prove it without having to reprove Strassen's theorem for $\mathbb{P}$? It doesn't look promising because the main connection of $\mathbb{P}$ and $\mu$ is having the same marginals (and that doesn't mean same measure eg. uniform on $[0,1]^{2}$ vs on diagonal in $[0,1]^{2}$).

This is because I am trying to prove $\mathbb{P}(d(X,Y) >\alpha +\varepsilon)\leq \varepsilon+\beta$. Any ideas for that(preferably only hints). Thanks

Attempt

We have $$\mathbb{P}(X\in A, Y\in B)-\mu(A\times B)=\mathbb{P}(X\in A)+\mathbb{P}( Y\in B)-\mathbb{P}(X\in A\cup Y\in B)-[\mu(A\times S)+\mu(S\times B)-\mu(A\times S\cup S\times B)]=\mu(A\times S\cup S\times B)-\mathbb{P}(X\in A\cup Y\in B)$$

Also,

$$\mu(d(x,y) < \alpha )=\mu(\bigcup_{x_{i}\in D} |x-x_{i}|< \alpha \times |y-x_{i}|< \alpha),$$

where D is the separable subset of X. I am trying to connect this with $\mathbb{P}(\bigcup_{x_{i}\in D} |X-x_{i}|< \alpha \cap |Y-x_{i}|< \alpha)$.

Given two random variables X,Y with measures P,Q. Show that if $P(E) \le Q(E^\alpha) + \beta$ for all measurable $E\subset\mathbb{R}$ then $\mathbb{P}(d(X,Y)>\alpha)<\beta$.

Only hints please.

Attempt

For all $\varepsilon>0$ Strassen's theorem gives us measure $\mu$ on $\mathbb{R}^{2}$ s.t. $\mu(d(X,Y) >\alpha +\varepsilon)\leq \varepsilon+\beta$ and marginals P,Q. Now I want to relate the measures $\mathbb{P}$ and $\mu$.

We have $$\mathbb{P}(X\in A, Y\in B)-\mu(A\times B)=\mathbb{P}(X\in A)+\mathbb{P}( Y\in B)-\mathbb{P}(X\in A\cup Y\in B)-[\mu(A\times S)+\mu(S\times B)-\mu(A\times S\cup S\times B)]=\mu(A\times S\cup S\times B)-\mathbb{P}(X\in A\cup Y\in B)$$

Also,

$$\mu(d(x,y) < \alpha )=\mu(\bigcup_{x_{i}\in D} |x-x_{i}|< \alpha \times |y-x_{i}|< \alpha),$$

where D is the separable subset of X. I am trying to connect this with $\mathbb{P}(\bigcup_{x_{i}\in D} |X-x_{i}|< \alpha \cap |Y-x_{i}|< \alpha)$.

Suppose that $(X,d)$ is a separable metric space and that $\alpha,\beta>0$. If $P$ and $Q$ are probability measures on $X$ satisfying $$ P(E) \le Q(E^\alpha) + \beta $$ for any Borel-measurable set $E \subset X$, then for any $\varepsilon>0$ there exist two nonnegative measures $\nu,\gamma$ on $X\times X$ such that

1. $\mu:=\gamma + \nu$ is a law on $X\times X$ with marginals $P$ and $Q$.

2. $\nu \{(x,y)\in X\times X: d(x,y) > \alpha + \varepsilon \} = 0$

3. $\gamma(X\times X) \le \beta + \varepsilon$

I want to prove that if real random variables X,Y on $(\Omega,\mathbb{P})$ have P,Q as distributions then

$$\mathbb{P}(d(X,Y) <\alpha )\geq \mu(d(x,y) < \alpha ).$$

Can I prove it without having to reprove Strassen's theorem for $\mathbb{P}$? It doesn't look promising because the main connection of $\mathbb{P}$ and $\mu$ is having the same marginals (and that doesn't mean same measure eg. uniform on $[0,1]^{2}$ vs on diagonal in $[0,1]^{2}$).

This is because I am trying to $\mathbb{P}(d(X,Y) >\alpha +\varepsilon)\leq \varepsilon+\beta$. Any ideas for that. Thanks

Attempt

We have $$\mathbb{P}(X\in A, Y\in B)-\mu(A\times B)=\mathbb{P}(X\in A)+\mathbb{P}( Y\in B)-\mathbb{P}(X\in A\cup Y\in B)-[\mu(A\times S)+\mu(S\times B)-\mu(A\times S\cup S\times B)]=\mu(A\times S\cup S\times B)-\mathbb{P}(X\in A\cup Y\in B)$$

Also,

$$\mu(d(x,y) < \alpha )=\mu(\bigcup_{x_{i}\in D} |x-x_{i}|< \alpha \times |y-x_{i}|< \alpha),$$

where D is the separable subset of X. I am trying to connect this with $\mathbb{P}(\bigcup_{x_{i}\in D} |X-x_{i}|< \alpha \cap |Y-x_{i}|< \alpha)$.

Suppose that $(X,d)$ is a separable metric space and that $\alpha,\beta>0$. If $P$ and $Q$ are probability measures on $X$ satisfying $$ P(E) \le Q(E^\alpha) + \beta $$ for any Borel-measurable set $E \subset X$, then for any $\varepsilon>0$ there exist two nonnegative measures $\nu,\gamma$ on $X\times X$ such that

1. $\mu:=\gamma + \nu$ is a law on $X\times X$ with marginals $P$ and $Q$.

2. $\nu \{(x,y)\in X\times X: d(x,y) > \alpha + \varepsilon \} = 0$

3. $\gamma(X\times X) \le \beta + \varepsilon$

I want to prove that if real random variables X,Y on $(\Omega,\mathbb{P})$ have P,Q as distributions then

$$\mathbb{P}(d(X,Y) <\alpha )\geq \mu(d(x,y) < \alpha ).$$

Can I prove it without having to reprove Strassen's theorem for $\mathbb{P}$? It doesn't look promising because the main connection of $\mathbb{P}$ and $\mu$ is having the same marginals (and that doesn't mean same measure eg. uniform on $[0,1]^{2}$ vs on diagonal in $[0,1]^{2}$).

This is because I am trying to prove $\mathbb{P}(d(X,Y) >\alpha +\varepsilon)\leq \varepsilon+\beta$. Any ideas for that(preferably only hints). Thanks

Attempt

We have $$\mathbb{P}(X\in A, Y\in B)-\mu(A\times B)=\mathbb{P}(X\in A)+\mathbb{P}( Y\in B)-\mathbb{P}(X\in A\cup Y\in B)-[\mu(A\times S)+\mu(S\times B)-\mu(A\times S\cup S\times B)]=\mu(A\times S\cup S\times B)-\mathbb{P}(X\in A\cup Y\in B)$$

Also,

$$\mu(d(x,y) < \alpha )=\mu(\bigcup_{x_{i}\in D} |x-x_{i}|< \alpha \times |y-x_{i}|< \alpha),$$

where D is the separable subset of X. I am trying to connect this with $\mathbb{P}(\bigcup_{x_{i}\in D} |X-x_{i}|< \alpha \cap |Y-x_{i}|< \alpha)$.