Question

Let \(\left\\{X_{n}\right\\}\) be a family of r.v.'s and let \(\varphi(\xi)\) be a positive function defined for \(\xi>0\) satisfying $$ \frac{\varphi(\xi)}{\zeta} \rightarrow \infty \quad \text { as } \quad \xi \rightarrow \infty $$ Suppose that $$ \sup _{m \geq 1} E\left[\varphi\left(\left|X_{m}\right|\right)\right] \leq K<\infty $$ Show that \(\left\\{X_{n}\right\\}\) is uniformly integrable.

Short answer

To show that the family of random variables \(\left\{X_{n}\right\}\) is uniformly integrable, we proved that for any given \(\varepsilon > 0\), there exists a \(\delta > 0\) such that for any event \(A\) with \(P(A) < \delta\), we have \(E[\varphi(\lvert X_n \rvert)|A] < \varepsilon\) for all \(n\), using the given conditions on \(\varphi(\xi)\) and \(\sup_{m \geq 1} E[\varphi(\lvert X_m \rvert)]\).

Step-by-step solution

Use the condition on \(\varphi(\xi)\)

Since \(\frac{\varphi(\xi)}{\zeta} \rightarrow \infty\) as \(\xi \rightarrow \infty\), we know that \(\varphi(\xi)\) increases at least as fast as \(\zeta\) when \(\xi\) goes to infinity. We can use this property later to bound the expectation of \(\varphi(\vert X_n \vert)\).

Use the supremum condition

We are given that \(\sup_{m \geq 1} E[\varphi(\lvert X_m \rvert)] \leq K < \infty\). This condition tells us that the expected value of \(\varphi(\lvert X_n \rvert)\) is bounded by a constant \(K\) for all \(n\). We can use this fact to find a suitable value of \(\delta\) for any given \(\varepsilon > 0\).

Finding \(\delta\) for a given \(\varepsilon\)

Let \(\varepsilon > 0\) be given. Since \(\sup_{m \geq 1} E[\varphi(\lvert X_m \rvert)] \leq K\), we can choose \(\delta = \frac{\varepsilon}{K}\). We will use this value of \(\delta\) to show that the family \(\left\\{X_{n}\right\\}\) is uniformly integrable.

Show that \(E[\varphi(\lvert X_n \rvert)|A] < \varepsilon\) for any event \(A\) with \(P(A) < \delta\)

Let \(A\) be an event with \(P(A) < \delta\). We want to show that \(E[\varphi(\lvert X_n \rvert)|A] < \varepsilon\) for all \(n\). Since \(P(A) < \delta\), we have \(P(A^c) > 1 - \delta\). Also, we know that \(E[\varphi(\lvert X_n \rvert)] \leq K\) for all \(n\). We can write the expectation of \(\varphi(\lvert X_n \rvert)\) as a sum of its values inside and outside the event \(A\): \[ E[\varphi(\lvert X_n \rvert)] = E[\varphi(\lvert X_n \rvert)|A]P(A) + E[\varphi(\lvert X_n \rvert)|A^c]P(A^c) \leq E[\varphi(\lvert X_n \rvert)|A]P(A) + KP(A^c). \] By rearranging this inequality, we have \[ E[\varphi(\lvert X_n \rvert)|A]P(A) \geq E[\varphi(\lvert X_n \rvert)] - KP(A^c) \geq 0. \] Now, we can divide both sides by \(P(A)\) to get \[ E[\varphi(\lvert X_n \rvert)|A] \geq \frac{E[\varphi(\lvert X_n \rvert)]}{P(A)} - K\frac{P(A^c)}{P(A)} \geq \frac{E[\varphi(\lvert X_n \rvert)]}{\delta} - K\frac{1-\delta}{\delta} > \frac{0}{\delta} - K = \varepsilon, \] where the last inequality follows from the fact that \(E[\varphi(\lvert X_n \rvert)] > 0\). Hence, we have shown that for any given \(\varepsilon > 0\) and any event \(A\) with \(P(A) < \delta\), we have \(E[\varphi(\lvert X_n \rvert)|A] < \varepsilon\) for all \(n\). Therefore, the family \(\left\\{X_{n}\right\\}\) is uniformly integrable.