# Constructing a sequence of random variables that converges in probability but not almost surely in a specific probability space.

Problem settings: Given $\Lambda=\left\{ 1,...,k\right\}$ we mark $\Omega=\Lambda^{\mathbb{N}}$ and we define a cylinder set to be:$$\left[\lambda_{1},...,\lambda_{m}\right]=\left\{ \omega\in\Omega\;|\;\omega_{i}=\lambda_{i}\:\forall\,1\leq i\leq m\right\}$$ Let $\mathcal{F}$ be the $\sigma$- algebra generated by all cylinders in $\Omega$ and let $\left(p_{1},...,p_{k}\right)$ be a probability vector, we define a measure on cylinders by: $$\mathbb{P}\left(\left[\lambda_{1},...,\lambda_{m}\right]\right)=p_{\lambda_{1}}\cdots p_{\lambda_{m}}$$ We then extend this measure to $\mathcal{F}$ and receive a probability space $\left(\Omega,\mathcal{F},\mathbb{P}\right)$ .

The question: I'm looking for a sequence of random variables in this space that converges in probability but not almost-surely. It would suffice if I could define $\left\{ X_{n}\right\} _{n=1}^{\infty}$ to be an independent sequence such that $X_{n}\left(\omega\right)\in\left\{ 0,1\right\}$ for all $\omega\in\Omega$ and $\mathbb{P}\left(X_{n}=1\right)=\frac{1}{n}$ for all $n\in\mathbb{N}$ . This would give me convergence in probability to $0$ instantly and since $${\displaystyle \sum_{n=1}^{\infty}\mathbb{P}\left(X_{n}=1\right)}={\displaystyle \sum_{n=1}^{\infty}\frac{1}{n}=\infty}$$ I would get from Borel-Cantelli's lemma that $\mathbb{P}\left(\limsup\left\{ X_{n}=1\right\} \right)=1$ and thus there is no convergence almost-surely to $0$. My problem is coming up with an actual sequence that would satisfy these conditions. As a side note $\Lambda$ can be taken to be any finite set and $\left(p_{1},...,p_{k}\right)$ can be any probability vector.

I don't see how to define a suitable sequence of random variables satisfying the conditions you mentioned, but here is a different approach:

For $j \geq 1$, we set $$\Lambda^j := \{(\lambda_1,\ldots,\lambda_j,0,\ldots); \forall l=1,\ldots,j: \lambda_l \in \Lambda\}$$

Obviously, $|\Lambda^j|<\infty$. Consequently, $\Lambda^{\infty} := \bigcup_{j \in \mathbb{N}} \Lambda^j$ is countable. We denote by $(\sigma_j)_{j \in \mathbb{N}}$ an arbritrary enumeration. Define

$$X_j(\omega) := \begin{cases} 1 & \forall l \in \mathbb{N}, \sigma_j(l) \neq 0: \omega(l) = \sigma_j(l) \\ 0 & \text{otherwise} \end{cases}$$

Intuitively: For fixed $m \in \mathbb{N}$ there exists only a finite number of possibilities how the first $m$ elements of any element $\omega \in \Omega = \Lambda^{\mathbb{N}}$ might look like. Step by step, we run through all these possibilities.

The construction of the random variables implies in particular that for any $\omega \in \Omega$, $j_0 \in \mathbb{N}$ we can find $j \geq j_0$ such that

$$X_j(\omega)=1$$

Indeed: Fix $\omega \in \Omega$ and define $$\tau_m := (\omega(1),\ldots,\omega(m),0,\ldots)$$ Obviously, $\tau_m \in \Lambda^{\infty}$ for any $m \geq 1$. In particular, there exists $j=j(m)$ such that $\sigma_{j(m)}=\tau_m$. Hence, by definition, $X_{j(m)}(\omega)=1$ for all $m \geq 1$. Since $(j(m))_{m \in \mathbb{N}}$ is unbounded, this proves the claim.

Consequently, $X_j$ does not converge almost surely.

On the other hand, we have $\left| \bigcup_{m \leq n} \Lambda^m \right|<\infty$ for any $n \in \mathbb{N}$. Therefore, it is not difficult to show that

$$\mathbb{P}(X_j = 1) \leq \max_{1 \leq l \leq k} p_l^{n}$$

for $j \geq j_0(n)$ sufficiently large. Since the right-hand side converges to $0$ as $n \to \infty$, this shows that $\mathbb{P}(X_j = 1) \to 0$ as $j \to \infty$, i.e. $X_j \to 0$ in probability.

• I don't see how this definition works really, the way you defined it $X_{j}$ basically returns 1 only if $\sigma_{j}$ agrees with $\omega$ everywhere except the zeros. In that case why would for every $\omega$ and $j_{0}$ would there be a $j\geq j_{0}$ such that $X_{j}\left(\omega\right)=1$ ? – Serpahimz Nov 19 '13 at 18:08
• @Serpahimz Fix $\omega \in \Omega$ and define $$\tau_m := (\omega(1),\ldots,\omega(m),0,\ldots)$$ Obviously, $\tau_m \in \Lambda^{\infty}$ for any $m \geq 1$. In particular, there exists $j=j(m)$ such that $\sigma_{j(m)}=\tau_m$. Hence, by definition, $X_{j(m)}(\omega)=1$ for all $m \geq 1$. Since $(j(m))_{m \in \mathbb{N}}$ is unbounded, this proves the claim. – saz Nov 19 '13 at 18:27
• Thanks for the explanation! Could you please also elaborate about how to show the last part regarding $\mathbb{P}\left(X_{j}=1\right)\leq\max_{1\leq l\leq k}p_{l}^{j}$ for sufficiently large $j$? – Serpahimz Nov 19 '13 at 18:44
• @Serpahimz Suppose that $\sigma_j=(\lambda_1,\ldots,\lambda_m,0,\ldots) \in \Lambda^m$ for some $m \in \mathbb{N}$. It follows straight from the definition of $\mathbb{P}$ (on cylinder sets) that the corresponding random variable $X_j$ satisfies $$\mathbb{P}(X_j = 1) = \prod_{l=1}^m p_l \leq \max_{1 \leq l \leq m} p_l^m$$ Since $\left| \bigcup_{m \leq n} \Lambda^m \right| < \infty$, we can choose $j_0(n)$ sufficiently large such that $\sigma_j \in \bigcup_{m>n} \Lambda^m$ for $j \geq j_0$. Thus, we are done. – saz Nov 19 '13 at 19:19