More convergence in probability, $L^p$ and almost surely
Let [math](\Omega,\A,\p)[/math] be a probability space. Let [math](X_n)_{n\geq 1}[/math] be a sequence of r.v.'s and assume that for all [math]\epsilon \gt 0[/math] we have
Take [math]\epsilon_k=\frac{1}{k}[/math] for [math]k\in\N[/math] with [math]k\geq 1[/math]. Now with the Borel-Cantelli lemma we get
Example
Let [math](\Omega,\A,\p)[/math] be a probability space. Let [math](X_n)_{n\geq 1}[/math] be a sequence of r.v.'s such that [math]\p[X_n=0]=1-\frac{1}{1+n^2}[/math] and [math]\p[X_n=1]=\frac{1}{1+n^2}[/math]. Then for all [math]\epsilon \gt 0[/math] we get [math]\p[\vert X_n\vert \gt \epsilon]=\p[X_n \gt \epsilon]=\frac{1}{1+n^2}[/math], so it follows
which implies that [math]\lim_{n\to\infty\atop a.s.}X_n=0.[/math]
Let [math](\Omega,\A,\p)[/math] be a probability space. Let [math](X_n)_{n\geq 1}[/math] be a sequence of r.v.'s. Then
Exercise.
Example
Let [math](Y_n)_{n\geq 1}[/math] be iid r.v.'s such that [math]\p[Y_n\leq X]=1-\frac{1}{1+X}[/math] for [math]X\geq 0[/math] and [math]n\geq 1[/math]. Take [math]X_n=\frac{Y_n}{n}[/math] and let [math]\epsilon \gt 0[/math]. Then
and thus [math]\lim_{n\to\infty\atop \p}X_n=0[/math]. Moreover, we have
but [math]\prod_{m\geq n}^\infty\left(1-\frac{1}{1+m\epsilon}\right)=0[/math]. Hence [math]\p[\sup_{m\geq n}\vert X_n\vert \gt \epsilon]\not\rightarrow 0[/math] as [math]n\to\infty[/math] and therefore [math](X_n)_{n\geq 1}[/math] doesn't converge a.s. to [math]X[/math].
Let [math](\Omega,\A,\p)[/math] be a probability space. Let [math](X_n)_{n\geq 1}[/math] be a sequence of r.v.'s. Then [math]\lim_{n\to\infty\atop \p}X_n=X[/math] if and only if for very subsequence of [math](X_n)_{n\geq 1}[/math], there exists a further subsequence which converges a.s.
If [math]\lim_{n\to\infty\atop\p}X_n=X[/math], then any of its subsequences also converge in probability. We already know that there exists a subsequence which converges a.s. Conversely, if [math]\lim_{n\to\infty\atop\p}X_n=X[/math], then there is an [math]\epsilon \gt 0[/math], some [math]n_k\in\N[/math] and a [math]\nu \gt 0[/math] such that for all [math]k\geq 1[/math] we get
and therefore we cannot extract a subsequence from [math](X_{n_k})_{k\geq 1}[/math] which would converge a.s.
Let [math](\Omega,\A,\p)[/math] be a probability space. Let [math](X_n)_{n\geq 1}[/math] be a sequence of r.v.'s and [math]g:\R\to\R[/math] a continuous map. Moreover, assume that [math]\lim_{n\to\infty\atop\p}X_n=X[/math]. Then
Any subsequence [math]g((X_{n_k})_{k\geq 1})[/math] and [math](X_{n_k})_{k\geq 1}[/math] converges in probability. So it follows that there exists a subsequence [math](X_{m_k})_{k\geq 1}[/math] of [math](X_{n_k})_{k\geq 1}[/math] such that
because [math]g[/math] is continuous. Now with the previous lemma we get that
Let [math](\Omega,\A,\p)[/math] be a probability space. Let [math](X_n)_{n\geq 1}[/math] and [math](Y_n)_{n\geq 1}[/math] be sequences of r.v.'s such that [math]\lim_{n\to\infty\atop \p}X_n=X[/math] and [math]\lim_{n\to\infty\atop\p} Y_n=Y[/math]. Then
- [math]\lim_{n\to\infty\atop\p} X_n+Y_n=X+Y[/math]
- [math]\lim_{n\to\infty\atop \p}X_n\cdot Y_n=X\cdot Y[/math]
We need to show both points.
- Let [math]\epsilon \gt 0[/math]. Then [math]\vert X_n-X\vert\leq \frac{\epsilon}{2}[/math] and [math]\vert Y_n-Y\vert\leq \frac{\epsilon}{2}[/math] implies that [math]\vert (X_n+Y_n)-(X+Y)\vert\leq \epsilon[/math], and thus we get
[[math]] \p[\vert X_n+Y_n-(X+Y)\vert \gt \epsilon]\leq \p\left[\vert X_n-X\vert \gt \frac{\epsilon}{2}\right]+\p\left[\vert Y_n-Y\vert \gt \frac{\epsilon}{2}\right]. [[/math]]
- We apply proposition 8.4 to the continuous map [math]g(X)=X^2[/math]. Hence we get
[[math]] 2X_nY_n=(X_n+Y_n)^2-X_n^2-Y_n^2. [[/math]]
General references
Moshayedi, Nima (2020). "Lectures on Probability Theory". arXiv:2010.16280 [math.PR].