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We have $2$ groups of random variables $X_1, X_2, .. X_T$ and $Y_1, Y_2, .. Y_T$. The first group variables are NOT independent, but the second group are.

We know that

$$ P \{X_1 > a \} < P \{Y_1 > a\} $$


$$ P \{X_{t+1} > b | X_{t} = c \} < P \{c \times Y_{t+1} > b\} $$

Then is this inequality

$$ P\{X_t > a\} < P\left\{\prod_{i=1}^tY_i < a\right\} $$


For me it seems to be correct intuitively. In a certain sense, $X_t$ is smaller than $\prod_{i=1}^tY_i$.

But can we prove it?

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up vote 2 down vote accepted

All kinds of order reversals can occur when products of negative real numbers enter the picture so let us assume that $X_t$ and $Y_t$ are nonnegative with full probability, for every $t$.

A key fact in this context is the following coupling result:

Assume that the random variables $\xi$ and $\eta$ are such that $\mathrm P(\xi\geqslant x)\leqslant\mathrm P(\eta\geqslant x)$ for every $x$. Then there exists random variables $\xi'$ and $\eta'$, defined on a common probability space, such that $\xi'$ is distributed like $\xi$, $\eta'$ is distributed like $\eta$, and $\xi'\leqslant\eta'$ with full probability.

Of course, the reciprocal holds since, if such random variables $\xi'$ and $\eta'$ exist, then, for every $x$, $[\xi'\geqslant x]\subseteq[\eta'\geqslant x]$, hence $$ \mathrm P(\xi\geqslant x)=\mathrm P(\xi'\geqslant x)\leqslant\mathrm P(\eta'\geqslant x)=\mathrm P(\eta\geqslant x). $$

Let us apply this key fact to show that $\mathrm P(X_t\geqslant x)\leqslant\mathrm P(U_t\geqslant x)$ for every $x\geqslant0$, recursively over $t\geqslant1$, where $U_t=Y_1Y_2\cdots Y_t$.

The case $t=1$ is part of the hypothesis.

Assume the result holds for some $t\geqslant1$. Then, introducing the distribution $\mu_t$ of $X_t$, $$ \mathrm P(X_{t+1}\geqslant x)=\mathrm E(\mathrm P(X_{t+1}\geqslant x\mid X_t))\leqslant\int\mathrm P(zY_{t+1}\geqslant x)\mu_t(\mathrm dz)=(*). $$ Let $X$ denote any random variable independent on $Y_{t+1}$ and distributed like $X_t$. Then, $$ (*)=\mathrm P(XY_{t+1}\geqslant x). $$ The key fact applied to the recursion hypothesis shows that there exists some random variables $X'$ and $U'$ such that $X'$ is distributed like $X$, $U'$ is distributed like $U_t$, and $U'\geqslant X'$ almost surely. Let $Y'$ denote any random variable independent on $(X',U')$ and distributed like $Y_{t+1}$. Then, $$ (*)=\mathrm P(X'Y'\geqslant x). $$ Since $Y'\geqslant0$ almost surely, $U'Y'\geqslant X'Y'$ almost surely and $[X'Y'\geqslant x]\subseteq[U'Y'\geqslant x]$. Thus, $$ (*)\leqslant\mathrm P(U'Y'\geqslant x). $$ Note that $(U',Y')$ is distributed like $(U_t,Y_{t+1})$ since $U'$ is independent on $Y'$, $U'$ is distributed like $U_t$ and $Y'$ is distributed like $Y_{t+1}$. Hence, $$ \mathrm P(U'Y'\geqslant x)=\mathrm P(U_tY_{t+1}\geqslant x)=\mathrm P(U_{t+1}\geqslant x), $$ which concludes the proof that $\mathrm P(X_{t+1}\geqslant x)\leqslant\mathrm P(U_{t+1}\geqslant x)$.

Edit 1: The OP asks for some explanations about the relation $(*)=\mathrm P(XY_{t+1}\geqslant x)$. This follows from the definitions and from the independence property. To see this, introduce the distribution $\nu_{t+1}$ of $Y_{t+1}$ and note that, by definition, $$ (*)=\int\mathrm P(zY_{t+1}\geqslant x)\mu_t(\mathrm dz)=\iint [zy\geqslant x]\mu_t(\mathrm dz)\nu_{t+1}(\mathrm dy), $$ that is, $$ (*)=\iint u(z,y)\mu_t(\mathrm dz)\nu_{t+1}(\mathrm dy),\quad\text{with}\ u:(z,y)\mapsto[zy\geqslant x]. $$ Let $(X'',Y'')$ denote any couple of random variables with distribution $\mu_t\otimes\nu_{t+1}$. In other words, assume that the distribution of $X''$ is $\mu_t$, the distribution of $Y''$ is $\nu_{t+1}$ and $X''$ and $Y''$ are independent. Then, $$ (*)=\mathrm E(u(X'',Y''))=\mathrm P(X''Y''\geqslant x). $$ To conclude, note that $(X,Y_{t+1})$ is an example of such a couple $(X'',Y'')$.

Edit 2: The coupling result mentioned at the beginning of this post is a consequence of the following fact, often called Skorokhod representation theorem (see the first chapter of The coupling method by T. Lindvall):

For every random variables $\xi$ and $\eta$, there exists a random variable $\zeta$, uniform on $(0,1)$, and some nondecreasing functions $u$ and $v$ such that $u(\zeta)$ is distributed like $\xi$ and $v(\zeta)$ is distributed like $\eta$.

Since, in our case, $\mathrm P(\xi\geqslant x)\leqslant\mathrm P(\eta\geqslant x)$ for every $x$, one can choose $u$ and $v$ such that $u\leqslant v$. Hence, a solution (a so-called coupling) is $\xi'=u(\zeta)$ and $\eta'=v(\zeta)$.

The basic idea of this version of Skorokhod representation theorem is to pick for $u$ the inverse of the CDF $F_\xi:x\mapsto\mathrm P(\xi\leqslant x)$ of $\xi$ and for $v$ the inverse of the CDF $F_\eta:x\mapsto\mathrm P(\eta\leqslant x)$ of $\eta$. Beware however that, to be fully rigorous, one must define carefully these so-called inverses since, in the general case, $F_\xi$ and $F_\eta$ need not be continuous nor strictly increasing (hence one relies on formulas like $u(z)=\inf\{x\mid F_\xi(x)\geqslant z\}$ and $v(z)=\inf{x\mid F_\eta(x)\geqslant z}$). Nevertheless, omitting the technical details of the representation, one can guess that $F_\xi\geqslant F_\eta$ implies $u\leqslant v$, which proves the result.

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Do you know where I can find more about the coupling result you referred to in your proof? This is the first time I see it. – ablmf Feb 26 '12 at 2:02
And how did you get $(*) = P(XY_{t+1} \ge x)$. Can you explain a bit more? – ablmf Feb 26 '12 at 2:44
On coupling: see Edit 2. About $(*)$: see Edit 1. – Did Feb 26 '12 at 9:52

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