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Consider a sequence of independent $0$-or-$1$ outcomes $X_1, \ldots, X_n,\ldots$, such that $\mathbb{P}(X=1)=p$. The discrete time process $\left(Y_n\right)_{n \geqslant 0}$, such that $$Y_n = \sum_{k=1}^n X_k$$ is called binomial process. Let $\tau_i$ denote ordinal numbers where $X_{\tau_i}=1$. These are event times of the binomial process. Letting $\tau_0=0$, the times between events, $T_i = \tau_{i}-\tau_{i-1}$ are independent and follow Pascal distribution: $$ \mathbb{P}\left(T_i = r\right) = p (1-p)^{r-1} [ r \geqslant 1 ] $$ Clearly $\tau_i < \tau_j$ for all $i<j$.

Q: For a fixed $n > 0$, given $Y_n = m > 0$, I am seeking to determine the distribution of event times $ (\tau_1, \ldots, \tau_m)$.

On other words, for a fixed $0 < m \leqslant n$, I am seeking to determine the distribution of $(T_1, T_1+T_2, \ldots, T_1+\cdots+T_m)$ given that $T_1+\cdots+T_m \leqslant n$ and $T_1+\cdots+T_m +T_{m+1} > n$.

The question is motivated by oft-drawn analogy of binomial process and Poisson process, and the nice corresponding result for the Poisson process, where interarrival times are exponentially distributed, and conditional on there being $m$ events on the interval $(0, \tau)$, arrival times are order statistics of uniform distribution on $(0,\tau)$. I am hoping some similarly neat result could be established for the case of binomial process as well.

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2 Answers 2

up vote 3 down vote accepted

Conditionally on $[Y_n=m]$ the set $T=\{\tau_k\mid1\leqslant k\leqslant m\}$ is uniformly distributed on the space $\mathfrak T_m$ of subsets of size $m$ of $\{1,2,\ldots,n\}$.

Proof: for every $t$ in $\mathfrak T_m$, the event $[T=t]$ has absolute probability $p^m(1-p)^{n-m}$. This probability does not depend on $t$ in $\mathfrak T_m$ hence the result follows. In particular, $$ \mathbb P(T=t\mid Y_n=m)=|\mathfrak T_m|^{-1}={n\choose m}^{-1}. $$ For example, the number of $t$ in $\mathfrak T_m$ such that $\min t=k$ is $$ {n-k\choose m-1}, $$ hence $$ \mathbb P(\tau_1=k\mid Y_n=m)={n-k\choose m-1}\cdot{n\choose m}^{-1}. $$

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+1 and Thanks! After seeing your answer, I also came up with more brute-force demonstration. –  Sasha Feb 28 '13 at 20:09

Although not nearly as elegant as @Did's proof, here is a brute-force one. Denote the event $\mathcal{E}_{n,m} = \{Y_n=m\}=\{ T_1+\cdots+T_m \leqslant n, T_1+\cdots+T_m + T_{m+1} > n \}$. Then we are after $$ \mathbb{P}\left(\tau_1=r_1,\tau_2=r_2,\ldots,\tau_m=r_m \mid \mathcal{E}_{n,m} \right) = \frac{ \mathbb{P}\left(T_1=r_1,T_2=r_2-r_1,\ldots,T_m=r_m-r_{m-1},T_{m+1}>n-r_m\right)}{ \mathbb{P}\left(\tau_m \leqslant n, \tau_m + T_{m+1} > n\right) } \tag{1} $$ Using the independence, the numerator reads: $$\begin{eqnarray} \mathbb{P}\left(T_{m+1}>n-r_m\right) \prod_{k=1}^m \mathbb{P}\left(T_k=r_k - r_{k-1}\right) &=& (1-p)^{n-r_m} \prod_{k=1}^m \left( p (1-p)^{r_k-r_{k-1}-1}\right) \\ &=& p^m (1-p)^{n-m} \end{eqnarray} $$ The denominator reads: $$ \begin{eqnarray} \sum_{k=m}^n \mathbb{P}\left(\tau_m=k\right) \mathbb{P}\left(T_{m+1}>n-k\right) &=& \sum_{k=m}^n \binom{k-1}{m-1} p^m (1-p)^{k-m} (1-p)^{n-k} \\ &=& p^m (1-p)^{n-m} \sum_{k=m}^n \binom{k-1}{m-1} \end{eqnarray} $$ Using $\binom{k-1}{m-1} = \binom{k}{m} - \binom{k-1}{m}$ the sum telescopes to give $\binom{n}{m}$.

Substituting expressions for the numerator and the denominator in eq. $(1)$ we recover @Did's result: $$ \mathbb{P}\left(\tau_1=r_1,\tau_2=r_2,\ldots,\tau_m=r_m \mid Y_n=m \right) = \frac{p^m (1-p)^{n-m}}{\binom{n}{m} p^m (1-p)^{n-m} } = \frac{1}{\binom{n}{m}} $$

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