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**Moo** Laurent : Now I've read your explanation... what is a cylinder subset ??

It is the kind of subsets that is used to define the product $\displaystyle \sigma$-algebra $\displaystyle \mathcal{A}^{\otimes \mathbb{N}}$ on $\displaystyle X^{\mathbb{N}}$, where $\displaystyle (X,\mathcal{A})$ is a measurable space. The cylinder subsets of $\displaystyle X^{\mathbb{N}}$ are the subsets like $\displaystyle \prod_{i=0}^n A_i \times X^{\mathbb{N}}=\{(x_i)_{i\geq 0}|x_0\in A_0,\, x_1\in A_1,\ldots x_n\in A_n\}$ or, in other words, the subsets defined by conditions on a finite number of coordinates. In the present case, $\displaystyle X=\{0,1\}$.

After that, I'm asked to show that $\displaystyle P(T_{r+1}-T_r=i ~,~ T_r=j)=P(S_i=1 ~,~ S_{i-1}=0)P(S_j=r ~,~ S_{j-1}=r-1)$

So for this question, I "translated" the RHS into $\displaystyle P(T_1=i)P(T_r=j)=(1-p)^{j-1} p \cdot P(T_r=j)$

And then I said that the LHS = $\displaystyle P(T_{r+1}-T_r=i \mid T_r=j)P(T_r=j)$$\displaystyle =P(X_{j+1}=\dots=X_{j+i-1}=0~,~X_{i+j}=1)P(T_r=j)$

Then it was clear that LHS=RHS.

However, is this a correct method ?

The equality with the conditional expectation is true but not clear; it almost assumes the next question as given. Instead, you can just describe the event and use independence (since $\displaystyle \{T_r=j\}$ depends on $\displaystyle X_1,\ldots,X_j\}$ only):

$\displaystyle P(T_r=j, T_{r+1}-T_r=i)=P(T_r=j, X_{j+1}=0, \ldots, X_{j+i-1}=0, X_{j+i}=1)= $ $\displaystyle P(T_r=j)P(X_{j+1}=0, \ldots, X_{j+i-1}=0, X_{j+i}=1)$ $\displaystyle =P(T_r=j)P(T_1=i)$

and the last equality is because $\displaystyle (X_{j+k})_{k\geq 1}$ is distributed like $\displaystyle (X_k)_{k\geq 1}$.

Thereafter, I have to deduce that $\displaystyle T_{r+1}-T_r$ and $\displaystyle T_r$ are independent.... If they are, it is obvious that we get the previous formula. But how to prove they're independent ??

This is almost obvious from the previous formula. Indeed, the equality $\displaystyle P(T_r=j,T_{r+1}-T_r=i)=P(T_r=j)P(T_1=i)$ shows that the distribution of the random variable $\displaystyle (T_r,T_{r+1}-T_r)$ is a product distribution.

You don't even have to say that $\displaystyle T_{r+1}-T_r$ and $\displaystyle T_1$ have the same distribution; this comes as a by-product (anyway, it was almost proved in the previous question).