0.1 Convergence in \(L_{p}\) Norm
Recall that a norm is any function \(\Vert \cdot\Vert\) from a vector (function) space \(V\) to \(\mathbb{R}\) satisfying:
- \(\Vert f\Vert \geq 0\) for every \(f\in V\)
- \(\Vert f\Vert=0\) if \(f\) is a zero function
- \(\Vert af\Vert = |a|\Vert f\Vert\) for every \(a \in \mathbb{R}\), \(f \in V\)
- \(\Vert f+g \Vert \leq\Vert f\Vert +\Vert g\Vert\) (triangle inequality)
The \(L^p\) norm is a specific norm intimately linked with probability spaces and expectations defined as \[ \Vert X\Vert_{p}=[\mathbb{E}|X|^p]^{1/p}=\left(\int _{\Omega}|X|^p \, d\mathbb{P}\right)^{1/p} . \] :::{.definition data-title=“Convergence in Lp Norm”}
A sequence of random variables \(\{ X_{n} \}_{n\in\mathbb{N}}\) convergence in \(L^p\)-norm to \(X\) denoted \(X_{n}\stackrel{L^p}{\to}X\) if \[\lim_{ n \to \infty }\Vert X_{n}-X\Vert_{p}=\lim_{ n \to \infty }(\mathbb{E}|X_{n}-X|^p)^{1/p} = 0.\]
:::
Random variables can also converge almost surely, in probability and in law and the relationships between all types of convergence can be found in the note: Convergence of Random Variables.
Figure: Random Variable Convergence Relationships.
0.2 Convergence in Norm & Convergence in Probability
The key results relating to convergence in norm are detailed below.
Consider \(1<r<s<\infty\) and assume that \(X_{n}\stackrel{L^s}{\to}X\) as \(n \to \infty\). Then we have that \(X_{n}\stackrel{L^r}{\to}X\). In other words, higher orders of r-th mean convergence imply lower orders.
Proof (1): This result follows directly from applying Lyapunov's Inequality \[ (\mathbb{E}|X|^r)^{1/r}\leq(\mathbb{E}|X|^s)^{1/s};\quad 0<r<s<\infty. \] Note: The converse does not hold in general.
Proof (2): The result holds by application of Jensen's Inequality using the convex function \(f(x)=x^{\frac{s}{r}}\).
If \(X_{n}\stackrel{L^1}{\to}X\) then \(X_{n}\stackrel{\mathbb{P}}{\to}X\) as \(n \to \infty\).
Proof: This result follows directly from applying Markov Inequality since, for every \(\epsilon>0\) we have that \[ \mathbb{P}(|X_{n}-X|>\epsilon)\leq \frac{1}{\epsilon}\cdot \mathbb{E}|X_{n}-X|. \] Note: The converse does not hold in general however with the additional constraint of uniform integrability it does.
For \(p>0\) assume that \(X_n \in L^p\), \(n \geq 1\) and \(X_{n}\stackrel{\mathbb{P}}{\to} X\). The following three statements are equivalent: 1. \(\{ |X_{n}|^p,~n\geq 1 \}\) is uniformly integrable; 2. \(X_{n}\stackrel{L^p}{\to}X\) and \(X\in L^p\); and 3. \(\lim_{n \to \infty}\mathbb{E}[|X_{n}|^p]=\mathbb{E}[|X|^p]<\infty\).