An Easier Way to Solve Poisson Processes Problem

Question

So the question is :

Assume that Mary needs to sell one share of an asset by posting a limit sell order. She always keeps the first position in the queue, so that the first market buy order will execute her limit sell order and will complete the task. Market orders arrive according to a Poisson process with rate $λ $ per second, and each of them is equally likely to be a buy or a sell order, independent of everything else. Assume also that Mary’s client is impatient and will cancel the task (i.e. force Mary to withdraw her limit order) in $T $ seconds. It is known that the first market order will be a sell order. What is the probability that Mary will complete her task before the client withdraws?

My solution: I want to compute $P(N_T^B \geq 1) $ which means by the time of $T$, there is at least one buy order.

Introduce the random variables {$X_i$}$_{i=1}^∞$, taking values in {$0,1$}, which indicate whether the i-th order was sell $(0)$ or buy $(1)$.

Since $X_i$’s are i.i.d. and

$P(X_i=1)= λ/(λ+λ) = 0.5$,

Then, $P(N_T^B \geq 1)= P(X_2=1∪X_3=1 ∪ ... ∪ X_{T-1}=1) = 0.5 $

Can I use this method to solve this problem? The similar method is used to solve another pois. process promblem - selling concert tickets on the book, but I am not sure if my solution to this problem correct.

If not, how can I compute $P(N_T^B \geq 1) $ ?

Update : Using law of complements

$1-P(n=0) = 1-e^{(-λT/2)}(λT)^n/n!$ where $n = 0$

$P(N_T^B \geq 1) $= $1-e^{(-λT/2)}$

Answer 1

The index of the sequence $\{X_i\}$ is not the time of arrival. $X_1$ is the first order, not the order at time 1.

You do not know how many orders arrive during time $T$.   The number of orders, $N_T$, is Poisson distributed.   So using the Law of Total Probability partitioned over number of arrivals.

$$\begin{align}\mathsf P(N_T^B\geq 1) & = 1-\mathsf P(N_T^B=0) \\ & = 1- \sum_{n=0}^\infty \mathsf P(N_T^B=0\mid N_T=n)\,\mathsf P(N_T=n) \\ & = 1-\mathsf P(N_T=0)-\sum_{n=1}^\infty \mathsf P(\bigcup_{i=1}^{n} \{X_i=0\})\,\mathsf P(N_T=n)\end{align}$$

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The other method is okay, and can be used to verify your answer obtained by using the indicator functions (as requested).

Answer 2

The probability we should compute here is $P(N_T^B \ge 1 | X_1=1)$, not $P(N_T^B \ge 1)$. And there is a mistake in @Graham answer: $$ P(N_T^B=0|N_T=n) \ne P(\cup{\{X_i=0\}}) $$