Example Problems on Convergence of Random Variables
ST779 Homework 10 on Convergence of Random Variables
1
Let Ω be a countable sample space with the power set P(Ω) as the ω-field on Ω (that is, all subsets of Ω are measurable). Let Xn, X be random variables on Ω, such that Xnp→X. Show that Xna.s.→X.
Take ˆω to be the set of singletons with nonzero probability. Then take ωi∈ˆω. Since Xnp→X, we know that Xn(ωi)p→X(ωi). And we know for ε>0, ∃N such that for n>N,
P(∣Xn(ωi)−X(ωi)∣>ε)<P(ωi)≤∑P(ωi)=1.Now we can use the Borel-Cantelli Lemma. Take An={ω:∣Xn(ω)−X(ω)∣<ε}. Then,
P(lim supThus, X_{n} \stackrel{ \text{a.s.}}{\rightarrow} X.
2
Show that X_{n} \stackrel{ \text{p}}{\rightarrow} X if and only if E \Big[ \Big( \mid X_{n} - X \Big) / \Big( 1+\mid X_{n} - X \mid \Big)\Big] \rightarrow 0.
Notice that
\frac{\Big( \mid X_{n} - X \mid \Big)}{ \Big( 1+\mid X_{n} - X \mid \Big)} \leq \mid X_{n} - X \mid. \RightarrowAssume P( \mid X_{n} - X \mid ) > \varepsilon) \leq \varepsilon. We can break the expected value into two pieces.
\begin{align} E & \Big[ \frac{ \Big( \mid X_{n} - X \Big)} { \Big( 1+\mid X_{n} - X \mid \Big)} \Big] \\ & = E \Big[ \frac{ \Big( \mid X_{n} - X \Big)}{\Big( 1+\mid X_{n} - X \mid \Big)} \mathbb{I}_{\mid X_{n} - X \mid ) < \varepsilon} \Big] + E \Big[ \frac{ \Big( \mid X_{n} - X \Big) }{ \Big( 1+\mid X_{n} - X \mid \Big)} \mathbb{I}_{\mid X_{n} - X \mid ) > \varepsilon} \Big] \\ & \leq E \Big[ \mid X_{n} - X \mid \mathbb{I}_{\mid X_{n} - X \mid ) < \varepsilon} \Big] + E \Big[ \mathbb{I}_{\mid X_{n} - X \mid ) > \varepsilon} \Big] \\ & \leq E \Big[ \varepsilon \mathbb{I}_{\mid X_{n} - X \mid ) < \varepsilon} \Big] + P(\mid X_{n - X} > \varepsilon) \\ & = \varepsilon + \varepsilon\\ & = 2 \varepsilon \end{align}Thus, E \Big[ \Big( \mid X_{n} - X \Big) / \Big( 1+\mid X_{n} - X \mid \Big)\Big] \rightarrow 0.
\LeftarrowNow assume E \Big[ \Big( \mid X_{n} - X \Big) / \Big( 1+\mid X_{n} - X \mid \Big)\Big] \rightarrow 0. Then we know for \varepsilon> 0, \exists N such that n > N, E \Big[ \Big( \mid X_{n} - X \Big) / \Big( 1+\mid X_{n} - X \mid \Big)\Big] \leq \varepsilon^2. Then, we know
\begin{align} \frac{ \varepsilon^{2} }{ 1 + \varepsilon } & \geq E \Big[ \frac{ \Big( \mid X_{n} - X \Big)}{\Big( 1+\mid X_{n} - X \mid \Big)}\Big] \\ & \geq E \Big[ \frac{ \varepsilon }{ 1 + \varepsilon } \mathbb{I}_{\mid X_[n] - X \mid < \varepsilon} \Big] \\ & = \frac{ \varepsilon }{ 1 + \varepsilon } P( \mid X_{n} - X \mid > \varepsilon) \\ \\ \Rightarrow \\ P( \mid X_{n} - X \mid > \varepsilon) & \leq \varepsilon. \end{align}Thus, X_{n} \stackrel{ \text{p}}{\rightarrow} X.
3
Fatou’s lemma for convergence in distribution: Let X_{n} be nonnegative random variables converging in distribution to X. Then show that E(X) \leq \liminf_{n \rightarrow \infty} E(X_{n}).
Since X_{n} \stackrel{ \text{d}}{\rightarrow} X, we know that P(X_{n} < x) \rightarrow P(X < x) pointwise. So we know
\lim_{n\rightarrow \infty} P(X_{n} < n) = \liminf_{n\rightarrow \infty} P(X_{n} < n) = P(X_{n} < n).Now we can apply Fatou’s Lemma.
E(X) = \int_{0}^{\infty} P(X \leq x) dx = \int_{0}^{\infty} \liminf P(X_{n} < x) dx \leq \liminf \int_{0}^{\infty} P(X_{n} \leq x) dx4
Let X_{n} be iid with E \Big( \mid X_{1} \mid^{p} \Big) < \infty, 0 < p < \infty. Show that n^{-1/p} \max \Big( \mid X_{1} \mid, \dots , \mid X_{n} \mid \Big) \stackrel{ \text{p}}{\rightarrow}0.
Take \varepsilon>0.
\begin{align} P & \Bigg(n^{-1/p} \max\Big( \mid X_{1}\mid ,\dots ,\mid X_{n} \mid \Big)\Bigg) > \varepsilon \Bigg) \\ & = P\Bigg(\max\Big( \mid X_{1}\mid ,\dots ,\mid X_{n} \mid \Big)\Bigg) > n^{1/p} \varepsilon \Bigg) \\ & = P\Bigg(\max\Big( \mid X_{1}\mid ,\dots ,\mid X_{n} \mid \Big)^{p} \Bigg) > n \varepsilon^{p} \Bigg) \\ & = 1 - P\Bigg(\max\Big( \mid X_{1}\mid ,\dots ,\mid X_{n} \mid \Big)^{p} \Bigg) \leq n \varepsilon^{p} \Bigg) \\ & = 1 - P\Bigg(\mid X_{1} \mid^{p} \leq n \varepsilon^{p} \Bigg)^{n} \end{align}Now we need to show P\Bigg(\mid X_{1} \mid^{p} \leq n \varepsilon^{p} \Bigg) \rightarrow 1. We can use Chebyshev’s inequality.
P\Bigg(\mid X_{1} \mid^{p} \leq n \varepsilon^{p} \Bigg) = 1 - P\Bigg(\mid X_{1} \mid^{p} > n \varepsilon^{p} \Bigg) \leq 1 - \frac{ E(\mid X_{1} \mid^{p}) }{ n \varepsilon^{p} } \rightarrow 1Thus,
n^{-1/p} \max \Big( \mid X_{1} \mid, \dots , \mid X_{n} \mid \Big) = 1 - P\Bigg(\mid X_{1} \mid^{p} \leq n \varepsilon^{p} \Bigg)^{n} \stackrel{ \text{p}}{\rightarrow} 1 - 1 = 0.5
Let X_{1}, X_{2}, \dots be iid standard exponential random variables. Show that \max \Big( X_{1} , \dots X_{n} \Big) - \log(n) converges in distribution. Find the limiting distribution as well.
Take Y_{n} = \max \Big( X_{1} , \dots X_{n} \Big) - \log(n). Then,
\begin{align} F_{Y_{n}}(y) & = P \Bigg( \max \Big( X_{1} , \dots X_{n} \Big) - \log(n) \leq y \Bigg) \\ & = P \Bigg( \max \Big( X_{1} , \dots X_{n} \Big) \leq y + \log(n)\Bigg) \\ & = \prod_{i=1}^{n} P(X_{i} \leq y + \log(n)) \\ & = \Big( 1 - \exp \Big[ -(y + \log(n)) \Big] \Big)^{n} \\ & = \Big( 1 - \frac{ 1 }{ n }\exp \Big[ -y \Big] \Big)^{n} \\ \\ \lim_{n\rightarrow \infty} F_{Y_{n}}(y) & = \exp\Big[\exp[-y]\Big] \end{align}