Use Chebyshev's inequality to show that if X and Y are two...

Use Chebyshev's inequality to show that if $X$ and $Y$ are two arbitrary random variables satisfying $E\left\{(X-Y)^{2}\right\}=0$, then we have $P(X=Y)=1$, namely $X$ and $Y$ are almost surely identical. [Hint:
$P(|X-Y|>\epsilon)=0$ for any $\epsilon>0 .]$

Key Concepts

Chebyshev's Inequality

Chebyshev's inequality provides a bound on the probability that a random variable deviates from its mean by a certain amount in terms of its variance. It is a fundamental tool in probability theory because it applies to any random variable with a finite variance, regardless of its underlying distribution. In this context, the inequality is used to bound the probability that the difference between two random variables exceeds an arbitrary positive threshold.

Mean Squared Difference

The mean squared difference refers to the expected value of the squared difference between two random variables, which, in many contexts, is interpreted as a measure of the variability between them. When the mean squared difference is zero, it implies that there is no variability between the random variables; essentially, they must be equal almost everywhere. This concept is equivalent to having zero variance for the difference, signaling that the difference is almost surely zero.

Almost Sure Equality

Almost sure equality is a concept in probability theory which states that two random variables are equal with probability one. In other words, the set of outcomes where the two variables differ has probability zero. This concept is stronger than mere equality in distribution because it asserts that the two variables coincide on the entire probability space except possibly on a set of measure zero.

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