I have a Problem about hitting times.
That's the following:
Let $A\subset E$ and the first passage time $T_A$ and the hitting time $H_A$. Define: $T_A =\inf\{n\geq 0;X_n \in A\}$ and $H_A =\inf\{n\geq 1;X_n\in A\}$
Let $H_y :=\inf\{n\geq1:X_n =y\}$. To show: $\forall n>0: \sum_{m=1}^{n}{P_x(H_y=m)p_{y,y}^{n-m}}=p_{x,y}^n$.
So I tried this way:
\begin{align*} \sum_{m=1}^{n}{P_x(H_y=m)p_{y,y}^{n-m}} &=\sum_{m=1}^{n}{P(X_{n\wedge H_y(=m)}=y|X_0=x)P(X_{n-m}=y|X_0=y)}\\ &=\sum_{m=1}^{n}{P(X_{m}=y|X_0=x)P(X_{n-m}=y|X_0=y)}\\ &=\sum_{m=1}^{n}{\frac{P(X_{m}=y,X_0=x)}{P(X_0=x)}\frac{P(X_{n-m}=y,X_0=y)}{P(X_0=y)}}. \end{align*}
Now I was thinking to use the properties of the conditional expectation but I don't see the good way. Make this sense?
Thanks
I'd suggest approaching this using a first step analysis and induction. If $n=1$, then the result holds just by rearranging the definitions. Now assume that the result holds for some $n \geq 1$. Then you have $$ \sum_{m=1}^{n+1} P_x(H_y = m)p_{yy}^{(n-m+1)} = \sum_{z}\sum_{m=1}^{n+1}\overbrace{P_x(H_y=m\,|\,X_1=z)}^{=P_z(H_y=m-1)} \overbrace{P_x(X_1=z)}^{=p_{xz}}p_{yy}^{(n-m+1)}\\ =\sum_z p_{xz}\sum_{m=0}^nP_z(H_y=m)p_{yy}^{(n-m)} = p_{xy}p_{yy}^{(n)} + \sum_{z\neq y} p_{xz}\sum_{m=1}^n P_z(H_y=m)p_{yy}^{(n-m)}\\ =p_{xy}p_{yy}^{(n)} + \sum_{z\neq y}p_{xz}p_{zy}^{(n)}= \sum_z p_{xz}p_{zy}^{(n)} = p_{xy}^{(n+1)} $$ Now, you're done by induction. (Do you see where we used the induction hypothesis?)