I am stuck on the question below and I am not sure if I am heading in the right direction. Will be grateful for any direction.
Let $\left(S_{n}\right)_{n \geq 0}$ be a simple symmetric random walk. (a) Suppose that $m, n, x, y$ are integers such that $n>m>0$, and $n, y$ have the same parity. Prove that $$ P_{0}\left(S_{m}=x \mid S_{n}=y\right)=P_{0}\left(S_{n-m}=y-x \mid S_{n}=y\right) $$ (b) Hence compute $E_{0}\left(S_{m}+S_{n-m} \mid S_{n}=y\right)$ and $E_{0}\left(S_{m} \mid S_{2 m}=2 x\right)$.
My attempt
I am stuck on how to fine $P_{0}\left(S_{m}=x \mid S_{n}=y\right)$ but this is what I did:
Le $\varepsilon_{1}, \ldots, \varepsilon_{n}$ be elements of $\{-1,+1\}$ such that $\varepsilon_{1}+\cdots+\varepsilon_{n}=x$. Then I can write $P_{0}\left(S_{m}=x \mid S_{n}=y\right)$ as:
$$P_{0}\left(\varepsilon_{1} +, \ldots, + \varepsilon_{m}=x \mid S_{n}=y\right)$$
$$=\frac{P_{0}\left(\varepsilon_{1} +, \ldots, + \varepsilon_{m}=x\right)}{P_{0}\left(S_{n}=y\right)}$$
From the notes of Markov-chain and random walks by Takis, using Lemma 16 p.83, I know that $$P_{0}\left(S_{n}=y\right)=\left(\begin{array}{c}n \\ (n+y) / 2\end{array}\right) 2^{-n}$$
And since each $\varepsilon_{i}$ should be uniformly distributed then $$P_{0}(\varepsilon_{1} +, \ldots, + \varepsilon_{m}=x ) = \frac{1}{2^m}$$
Therefore: $$P_{0}\left(S_{m}=x \mid S_{n}=y\right) = \frac{2^{-m}}{\left(\begin{array}{c}n \\ (n+y) / 2\end{array}\right)2^{-n}}$$
Similarly:
$$P_{0}\left(S_{n-m}=y-x \mid S_{n}=y\right) =\frac{P_{0}\left(\varepsilon_{m} +, \ldots, + \varepsilon_{n}=y-x\right)}{P_{0}\left(S_{n}=y\right)} = \frac{2^{-(n-m)}}{\left(\begin{array}{c}n \\ (n+y) / 2\end{array}\right)2^{-n}}$$
I am not sure if what I am doing is correct so any insight would help me a lot.
For the second part where I need to find $E_{0}\left(S_{m}+S_{n-m} \mid S_{n}=y\right)$, I believe I can split this into $$E_{0}\left(S_{m} \mid S_{n}=y\right) + E_{0}\left(S_{n-m} \mid S_{n}=y\right)$$
Then I just take the expectation of the binomial distributed part and uniformly distributed part, am I right?
Thank you