Let $X_0,X_1,X_2,...$ be independent identically distributed nonnegative random variables having a continuous distribution. Let $N$ be the first index $k$ for which $X_k>X_0$. That is, $N=1$ if $X_1>X_0, N=2$ if $X_1≤X_0$ and $X_2>X_0$,etc.
Determine the probability mass function for $N$ and the mean $\mathbb{E}N$.
Interpretation: $X_0,X_1,X_2,...$ are successive offers or bids on a car that you are trying to sell. Then $N$ is the index of the first bid that is better than the initial bid.
This should not be too complicated, however I tried some different approaches with these conditional probabilities and I am really stuck on this task. I do not know how to handle it since the exact distribution of the random variables is not given. It appears that I have too little information.
Any kind of help or advice will be really appreciated.
Too long for a comment.
Since $X_0, X_1, X_2, \ldots$ are iid, given $X_0$, $X_i-X_0$ will be independent of $X_j-X_0$ for non-zero $i$ and $j$ s.t. $i\neq j$. Also, suppose that $F(x)$ is the cdf of $X_i$.
$$ \begin{align} P\{N = i|X_0=x\} &= P\{X_i>X_0, (X_{j}\leq X_0; 0 < j < i)|X_0=x\}\\ &= P\{X_i>X_0|X_0=x\}\prod_{j=1}^{i-1}P\{X_j \leq X_0|X_0=x\}\\ &= (1-F(x))F(x)^{i-1} \end{align} $$
$$\mathbb{E}(N|X_0=x) = \sum_{i=1}^{\infty}i(1-F(x))F(x)^{i-1} = (1-F(x))\frac{1}{(1-F(x))^2} = \frac{1}{1-F(x)}$$
$$\mathbb{E}N = \mathbb{E}(\mathbb{E}(N|X_{0})) = \mathbb{E}\left(\frac{1}{1-F(X_0)}\right) = \int_{0}^{\infty}\frac{f(x)}{1-F(x)}dx \rightarrow \infty$$
Edit:
$$ \begin{align} \mathbb{P}\{N=i\} &= \int_{0}^{\infty}\mathbb{P}\{N=i|X_0=x\}f(x)dx\\ &= \int_{0}^{\infty}(1-F(x))F(x)^{i-1}f(x)dx\\ &= \left(\frac{F(x)^i}{i}-\frac{F(x)^{i+1}}{i+1}\right)\bigg\vert_{0}^{\infty}\\ &= \frac{1}{i(i+1)} \end{align} $$
Verifying $\mathbb{E}N$,
$$\mathbb{E}N = \sum_{i=1}^{\infty}\frac{i}{i(i+1)} = \sum_{i=1}^{\infty}\frac{1}{i+1} \rightarrow \infty$$