Proof X has finite expectation and |E(X)−E(Y)|<M

Question

Suppose $X$ and $Y$ are two random variables such that $P (|X − Y | < M ) = 1$ for some constant $M$ . Show that if $Y$ has finite expectation, then $X$ has finite expectation and $|E(X) − E(Y) | < M$ .

Answer 1

Since $P(|X - Y| < M) = 1$ we can write

$$E[|X - Y|] = \int_{\Omega} |X - Y| dP = \int_{\{|X - Y| < M\}} |X - Y| dP + \int_{\{|X - Y| \geq M\}} |X - Y| dP$$

Since $P(|X - Y| < M) = 1$ we can say $\int_{\{|X - Y| \geq M\}} |X - Y| dP = 0$, and $\int_{\{|X - Y| < M\}} |X - Y| dP = \int_\Omega |X - Y| \mathbb{1}_{\{|X - Y| < M\}} dP < M P(|X - Y| < M) = M$.

In short, $E[|X - Y|] < M$.

Now using the fact that $E[|Y|] = C < \infty$, use the triangle inequality to get

$$E[|X|] = E[|X - Y + Y|] \leq E[|X - Y|] + E[|Y|] < M + C < \infty$$

Using the usual integral inequality we then get

$$|E[X] - E[Y]| = |E[X - Y]| = \left|\int_\Omega (X - Y) dP\right| \leq \int_\Omega |X - Y| dP = E[|X - Y|] < M$$