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Consider a symmetric random walk $P(X_i=1)=P(X_i=-1)=1/2$, $S_0=0$, $T_a=\min(n:S_n=a)$

We already know that $P(T_a>T_{-b})=1-P(T_{-b}< T_a)=\frac{b}{a+b}$ and $E(\min\{T_a,T_{-b}\})=ab$.

Prove: $$E(T_a \times 1\{T_a< T_{-b}\})=\frac{(a+2b)ab}{3(a+b)}.$$

I can rewrite the expectation as

$$E(\min\{T_a,T_{-b}\})=E(T_a \times 1\{T_a< T_{-b}\})+E(T_{-b} \times 1\{T_{-b}< T_a\})$$

But I don't know what to do next step? Anyone could give some hints? Thanks a lot.

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I have tried to format your question using $LaTeX$ to make it readable but you might want to check this is what you meant. –  Henry Jun 9 '12 at 9:54
    
Thanks a lot. This is the question i want to ask. –  Bonnie Jun 9 '12 at 10:23

1 Answer 1

up vote 1 down vote accepted

For every integer $-b\leqslant x\leqslant a$, let $t_x=\mathrm E_x(T_a:T_a\lt T_{-b})$. Then $t_{-b}=t_a=0$ and the Markov property after one step yields $$ t_x=\mathrm P_x(T_a\lt T_{-b})+\frac12(t_{x-1}+t_{x+1}), $$ for every integer $-b\lt x\lt a$. Write this as $$ (\Delta t)_x=-\mathrm P_x(T_a\lt T_{-b})=-\frac{x+b}{a+b}, $$ where $\Delta$ is the discrete Laplacian operator, defined by $$ (\Delta u)_x=\frac12(u_{x-1}+u_{x+1})-u_x. $$ Let us check the effect of $\Delta$ on some simple sequences:

  • If $u_x=1$ for every $x$, then $(\Delta u)_x=0$.
  • If $u_x=x$ for every $x$, then $(\Delta u)_x=0$.
  • If $u_x=x^2$ for every $x$, then $(\Delta u)_x=1$.
  • If $u_x=x^3$ for every $x$, then $(\Delta u)_x=3x$.

One sees that $\Delta t=\Delta t^{c,d}$, where, for every $(c,d)$, $t^{c,d}$ is defined by $$ t_x^{c,d}=\frac{c+dx-3bx^2-x^3}{3(a+b)}. $$ If $t_{-b}^{c,d}=t_a^{c,d}=0$, then $t=t^{c,d}$ on $\{-b,a\}$ and $\Delta t=\Delta t^{c,d}$ on $(-b,a)$, thus $\Delta (t-t^{c,d})=0$. The maximum principle shows that $t=t^{c,d}$ everywhere.

In particular, $t_0=\frac{c}{3(a+b)}$ if $(c,d)$ solves the system $t^{c,d}_{-b}=t^{c,d}_a=0$. This yields $c=ab(a+2b)$. QED.

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Thanks. I'm still a little confused. What's the meaning of $E_x$? –  Bonnie Jun 9 '12 at 21:58
    
$E_x=E(\ \mid S_0=x)$. –  Did Jun 10 '12 at 7:39

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