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If $X$ and $Y$ are independent continuous random variables with densities $f(x)$ and $g(y)$ then the probability that $Y$ is less than or equal to $X$ is

$$\Pr (Y \le X) = \int _{x=-\infty}^{\infty}(f(x)\int _{y=-\infty}^{x}g(y)dy)dx$$

Say $f(x)$ and $g(y)$ are parameterized with $n$ and we have $Var(X)$ and $Var(Y)$ converging to zero as $n\to\infty$, where at all times $E[Y]-E[X]>\epsilon$ for some $\epsilon>0$. How do we prove that, $$\Pr (Y \le X)\to0 $$ as $n\to\infty?$

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As stated, it is false. Don't forget that $\mathbb{E} (Y) - \mathbb{E} (X)$ may converge to $0$. – D. Thomine Jul 4 '12 at 11:00
thanks, I have made a small correction now. – Omri Jul 4 '12 at 11:16
up vote 1 down vote accepted

Motto: Forget the densities and come back to events.

  1. When $n\to\infty$, $\mathrm P(X_n\gt\mathrm E(X_n)+\frac12\epsilon)\to0$. (Hint: use Bienaymé-Chebychev inequality and the hypothesis that $\mathrm{Var}(X_n)\to0$.)
  2. When $n\to\infty$, $\mathrm P(Y_n\lt\mathrm E(Y_n)-\frac12\epsilon)\to0$. (Hint: use Bienaymé-Chebychev inequality and the hypothesis that $\mathrm{Var}(Y_n)\to0$.)
  3. For every $n$, $[Y_n\leqslant X_n]\subseteq[X_n\gt\mathrm E(X_n)+\frac12\epsilon]\cup[Y_n\lt\mathrm E(Y_n)-\frac12\epsilon]$.

Hence $\mathrm P(Y_n\leqslant X_n)\to0$ when $n\to\infty$.

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Thanks. Note that I need to use the one-sided Chebychev here (Cantelli's inequality)... – Omri Jul 5 '12 at 7:53
Omri: This is not true, (Bienaymé-)Chebychev (two-sided) inequality does the job perfectly well. – Did Jul 5 '12 at 8:19
Not sure I understand -- P(Xn-E(Xn) > t) is a one-sided probability, whereas P(|Xn-E(Xn)| > t) is the normal double-sided Chebychev... – Omri Jul 5 '12 at 9:24
Sure, and the first event is included in the second event hence any upper bound on the probability of the second event (provided by Bienaymé-Chebychev inequality) is also an upper bound on the probability of the first event. – Did Jul 5 '12 at 11:40

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