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2

The correct initial conditions are $p_0=1$ and $f_0=0$. The "1" is the first term $p_0 x^0$ which has to be separated out since the convolution formula (3) is only valid for $n\geq 1$. The rest is simple algebra: \begin{eqnarray*} P(x)&=&\sum_{n=0}^\infty p_n x^n\\[5pt] &=&1+\sum_{n=1}^\infty p_n x^n\\[5pt] &=&1+\sum_{n=1}^\infty \...

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For simplicity, I will write $q_k$ for $P(S=k)$ and $p_k$ for $P(X=k)$ while dropping the subscripts on $S_0$ and $X_1$. Your double summation is not quite correct; when you exchange the order of summation the lower bound should be $i=(k-1)_+$, not $i=k-1$. This is because $i$ can never be negative. Of course, this only makes a difference when $k=0$. The ...

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Your problem is equivalent to toss a $2k+1$ facets die at each step, subtract $k+1$ and get the result as the $\Delta x$ to move. Equivalently you can toss a die, with $2k+1$ facets, numbered $0,\; \ldots ,\,r=2k$, and subtract $k$. Let's take this model (it simplifies the treatment). So, you are asking what is the probability that after $m$ tosses you'd get ...

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Let $q$ be the probability that starting from $0$ the walk ever reaches $-1$. Then the probability $a_n$ that starting from $0$ the walk ever reaches $-n$ satisfies $$a_{n+1}=qa_n\;,$$ since the walk reaches $-(n+1)$ if and only if it first reaches $-n$ and then with probability $q$ reaches $-(n+1)$. Thus $a_n=q^n$. We have $q\ne1$, so the desired ...

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For the loss of $k$ to kick in, he needs to win $k-a$ times. If he does that, he will never go broke (except maybe on the round he would quit because of the $k$ losses). He needs to win those $k-a$ within the first $2k-a$ games. So compute the chance he goes broke in less than $2k-a$ games and the expected length of a game in that scenario. This gives ...

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Use conditional probability. For any positive integer $i$, and let $A_i$ be the random variable that takes the value of $1$ if and only if the $i$-th edge traversed in the roundtrip from $o$, if any, is $(x \to y)$. Then clearly $j_o(x \to y) = E[\sum_{1 \le i} A_i] = \sum_{1 \le i} E[A_i]$ (by linearity of expectation). Similarly, let $B_i$ be the random ...

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