Problem: Let $\lbrace X_k\rbrace_k$ be a sequence of independent random variables such that

\[E[X_k]=0, E[X_k^2]=1\]

Prove that $\lbrace\sum_{k=1}^n\frac{X_k}{\sqrt{n}}\rbrace_n$ are uniform integrable.

Solution: Lately, I’ve been thinking that learning mathematics is quite similar to cooking (I hope I can become a mathematician who also knows how to cook in the future). Let’s start solving this problem by preparing our ingredients—we need the following equivalent definition of uniform integrability:

Ingredient (Definition): $\lbrace X_k\rbrace_k$ are uniform integrable iff

\[\lim_{t\rightarrow 0}\sup_{k\geq 0}E[|X_k|\chi_{\lbrace|X_k|\geq t\rbrace}]=0\]

We now can have a try of an appetizer:

Appetizer (Lemma): If there exists some $p>1$ such that

\[\sup_{k\geq 0}E[|X_k|^p]<\infty\]

then $\lbrace X_k\rbrace_k$ are uniform integrable.

Proof: Let $q$ be the conjugate of $p$ ($p^{-1}+q^{-1}=1$). We use both Hölder’s and Chebyshev’s inequalities to deduce that

\[\begin{align*} E[|X_k|\chi_{\lbrace|X_k|\geq t\rbrace}]&\leq (E[|X_k|^p])^{1/p}(P(|X_k|\geq t))^{1/q}\\ &\leq M^{1/p}(P(|X_k|\geq t))^{1/q}\\ &\leq M^{1/p}(\frac{E[|X_k|^p]}{t^p})^{1/q}\\ &\leq \frac{M}{t^{p-1}} \end{align*}\]

Since $p>1$, we conclude that

\[\lim_{t\rightarrow 0}\sup_{k\geq 0}E[|X_k|\chi_{\lbrace|X_k|\geq t\rbrace}]\leq \lim_{t\rightarrow 0}\frac{M}{t^{p-1}}=0.\hspace{1mm}\square\]

Main Course (Proof of the Original Problem): By the lemma above, it suffices to prove that

\[\sup_{n\geq 0}E[|\frac{\sum_{k=1}^nX_k}{\sqrt{n}}|^2]<\infty\]

Let’s do some calculations!

\[\begin{align*} E[|\frac{\sum_{k=1}^nX_k}{\sqrt{n}}|^2]&=\frac{1}{n}E[(\sum_{k=1}^nX_k)^2]\\ &=\frac{1}{n}(\sum_{i\ne j}E[X_iX_j]+\sum_{i=1}^nE[X_i^2])\\ &=\frac{1}{n}(\sum_{i\ne j}E[X_i]E[X_j]+n)\\ &=1 \end{align*}\]

where we used the assumptions about mean and variance in the third equality. $\square$


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