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--- world:random-walk [2020/04/19 17:15]
rdouc [Proof]
+++ world:random-walk [2022/03/16 07:40]
127.0.0.1 external edit
@@ Line 4: / Line 4: @@
 <WRAP center round box 80%>
-Let $(U_i)$ be iid  Rademacher random variables, i.e. $U_i=1$ or $-1$ with probability $1/2$ and set $S_i=\sum_{j=1}^iU_j$ the associated partial sum. Define $\Delta=\inf\set{t>0}{S_t=0}$. What is the law of $\Delta$?
+Let $(U_i)$ be iid  Rademacher random variables, i.e. $U_i=1$ or $-1$ with probability $1/2$ and set $S_i=\sum_{j=1}^iU_j$ the associated partial sum. Define $\Delta=\inf\set{t>0}{S_t=0}$. Show that $S_n$ returns to 0 with probability one. What is the law of $\Delta$?
 </WRAP>
 ====== Proof ======
-Obviously, $\PP(\Delta=2k+1)=0$.
+Obviously, $\PP(\Delta=2k+1)=0$. It will be useful in some parts of the proof to note that, by symmetry, $\PP(\Delta=2k, S_1>0)=\PP(\Delta=2k, S_1<0)$, so that $\PP(\Delta=2k)=2\PP(\Delta=2k, S_1>0)$
 Define
 \begin{align*}
@@ Line 34: / Line 34: @@
 \frac{A(z)-z/4}{z/4}=A(z)\lr{1+\frac{A(z)-z/4}{z/4}}
 $$
-This is equivalent to $0=A^2(z)-A(z)+\frac{z}{4}$ so that $A(z)=\frac{1\pm \sqrt{1-z}}{2}$. Only one of the roots has an expansion with only non-negative coefficients, that is, $A(z)=\frac{1-\sqrt{1-z}}{2}$. Expanding the expression yields: $A(z)=\sum_{k=1}^{\infty} \begin{pmatrix}2k-2\\k-1\end{pmatrix}
+This is equivalent to $0=A^2(z)-A(z)+\frac{z}{4}$ so that $A(z)=\frac{1\pm \sqrt{1-z}}{2}$. Only one of the roots has an expansion with only non-negative coefficients, that is, $A(z)=\frac{1-\sqrt{1-z}}{2}$. This implies that
+$$
+\PP(\Delta<\infty)=\sum_{k=1}^\infty \underbrace{\PP(\Delta=2k)}_{2\PP(\Delta=2k,S_1>0)}= 2 A(1)=1
+$$
+that is, with probability 1, this random walk returns to 0.
+Moroever, expanding $\frac{1-\sqrt{1-z}}{2}$ yields: $A(z)=\sum_{k=1}^{\infty} \begin{pmatrix}2k-2\\k-1\end{pmatrix}
 \frac{1}{k 4^k} z^k$. Therefore, by symmetry, $\PP(\Delta=2k)=2\PP(\Delta=2k,S_1>0)=2\begin{pmatrix}2k-2\\k-1\end{pmatrix}\frac{1}{k 4^k}$.
-In particular, this implies that $\PP(\Delta<\infty)=\sum_{k=1}^\infty \PP(\Delta=2k)= 2 A(1)=1$: with probability 1, this random walk returns to 0.
-====== Comments ======
+====== Other method ======
-Maybe there exists a more simple way to find the expression of $\alpha_k$... A particular trajectory has a probability $(1/2)^{2k}=1/4^k$ to occur. It remains to count the number of trajectories. Since $U_1=1$ and $U_{2k}=-1$, it remains to count the number of trajectories of size $2k-2$ such that $S_i\geq 0$ for all intermediate time $i$ and $S_{2k-2}=0$. Without the constraint on the positivity of $S_i$ at each intermediate times, it would have been as if we take $k-1$ among $2k-2$ terms but in that case we count all the trajectories (even those that cross the $X$-axis), that is $\begin{pmatrix}2k-2\\ k-1\end{pmatrix}$. Comparing with the expression we finally obtained, we have find a justification for dividing this number by $k$. An idea would be to split them into well chosen equivalent classes of size $k$ with only one representant in each class with positive trajectories but I can't find this equivalence relation...
+Note that, setting $U_i$ such that $X_i=2U_i-1$, we have that $(U_i)$ are iid Bernoulli random variables with success probability $1/2$. Then,   $\PP(S_{2n}=0)=\PP(\sum_{i=1}^{2n} U_i=n)=\frac{(2n)!}{(n!)^2} \lr{\frac14}^{2n} \sim \frac{1}{\sqrt{n\pi}}$ where we have used the Stirling equivalence. This implies that
+\begin{equation}
+\label{eq:one}
+\infty=\sum_{n=1}^\infty \PP(S_{2n}=0)=\PE[\sum_{n=1}^\infty \indi{0}(S_{2n})]=\PE[\sum_{k=1}^\infty \indi{T^k<\infty}]=\sum_{k=1}^\infty \PP(T^k<\infty)
+\end{equation}
+where $T^k$ is the time index of the $k$-th visit of $(S_n)$ to $0$. By convention, we set $T^1=T$. We now show by induction that for all $k\geq 1$,
+\begin{equation} \label{eq:induc}
+\PP(T^k<\infty)=\PP(T<\infty)^k\eqsp.
+\end{equation}
+The case $k=1$ obviously holds. Now, assume that \eqref{eq:induc} holds for some $k\geq 1$. Then,
+\begin{align*}
+\PP(T^{k+1}<\infty)&=\PP(T^{k}<\infty,T^{k+1}<\infty)=\sum_{m=1}^\infty \sum_{n=1}^\infty\PP(T^{k}=m,T^{k+1}=m+n)\\
+&=\sum_{m=1}^\infty \sum_{n=1}^\infty\PP(T^{k}=m,\forall t\in[1:n-1],\ \sum_{\ell=1}^{t}X_{m+\ell}\neq 0,\sum_{\ell=1}^{n}X_{m+\ell}=0)\\
+&=\sum_{m=1}^\infty \sum_{n=1}^\infty\PP(T^{k}=m)\underbrace{\PP(\forall t\in[1:n-1],\ \sum_{\ell=1}^{t}X_{m+\ell}\neq 0,\sum_{\ell=1}^{n}X_{m+\ell}=0)}_{\PP(T=n)}\\
+&=\PP(T^k<\infty) \PP(T<\infty)
+\end{align*}
+and by the induction assumption, we get $\PP(T^{k+1}<\infty)=\PP(T<\infty)^{k+1}$.
+Plugging \eqref{eq:induc} into \eqref{eq:one} yields
+$$
+\infty=\sum_{k=1}^\infty \PP(T<\infty)^k
+$$
+Since $\PP(T<\infty) \in[0,1]$, this implies that $\PP(T<\infty)=1$.

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