How do I build a probability distribution from a $P(A_k| \cap_{i=1}^{k-1}A_i)=\theta^k$?

Question

A car windshield gets progressively weaker as it suffers from repeated debris strikes. Let $A_i$ be the event that the windshield survives the $i$th strike. For $0 < \theta < 1,$ let $P(A_1)=\theta$ and $P(A_k| \cap_{i=1}^{k-1}A_i)=\theta^k,$ where $k=2,3,\dots$. Define $X$ to be the number of debris strikes needed to break the windshield. Find the probability distribution of $X$.

I tried to obtain the probability distribution by going through the first few examples.

My work: $P(X=1)=(1-\theta), P(X=2)=\theta(1-\theta), P(X=3)=\theta^2(1-\theta),\dots$ So, $P(X=x)=\theta^x(1-\theta)$. However, I do not know if this work is correct, since the probability of the $i$th debris strike breaking the windshield may not necessarily be $(1-\theta)$.

Alternative work:

$P(A_1 \cap A_2 \cap \dots \cap A_{x-1} \cap A_x^c) = P(A_x^c | \cap_{i=1}^{x-1}A_i)P(\cap_{i=1}^{x-1}A_i)$

$=[1-P(A_x | \cap_{i=1}^{x-1}A_i)]P(A_{x-1}| \cap_{i=1}^{x-2}A_i)\dots P(A_2| A_1)P(A_1)$

$=(1-\theta^x)\theta^{x-1}\theta^{x-2}\dots \theta^2\theta =(1 - \theta^x)\theta^{(x-1)x/2}.$

So, $P(X=x)=p(x)=(1-\theta^x)\theta^{(x-1)x/2}$.

These two different approaches seem to yield different answers. Where am I messing up?

Answer 1

First I thought that the second way seems to be the flawed one but right now I am thinking the opposite: Of course, for general sets $A,B$ we do not necessarily have $P(A^C|B) = 1-P(A|B)$ however, in this case, this actually seems to be true. Reason:

$$\Omega = (A \cap B) ~\dot{\cup}~ (A^C \cap B) ~\dot{\cup}~ (A \cap B^C) ~\dot{\cup}~ (A^C \cap B^C)$$

so that

$$1 = P(\Omega) = P(A \cap B) + P(A^C \cap B) + P(A \cap B^C) + P(A^C \cap B^C)$$

In the case described above, $A = A_n$ and $B= \bigcap_{i=1}^{n-1} A_i$ so that $B^C \subset A^C$ (if the windshield was broken in one of the first debris strikes then it cannot survive the $n$-th strike either!). That has two important consequences:

1. $A^C \cap B^C = B^C$
2. $A \cap B^C = \emptyset$

i.e. the equation above simplifies to

$$1 = P(\Omega) = P(A \cap B) + P(A^C \cap B) + P(B^C)$$

now we divide by $P(B)$ to obtain

$$\frac{1}{P(B)} = \frac{P(A \cap B)}{P(B)} + \frac{P(A^C \cap B)}{P(B)} + \frac{P(B^C)}{P(B)}$$

The last summand is nothing else but

$$\frac{P(B^C)}{P(B)} = \frac{1-P(B)}{P(B)} = \frac{1}{P(B)} - 1$$

so that the equation further simplifies to

$$\frac{1}{P(B)} = \frac{P(A \cap B)}{P(B)} + \frac{P(A^C \cap B)}{P(B)} + \frac{1}{P(B)} - 1$$

and that indeed means nothing else but

$$1 = P(A|B) + P(A^C|B)$$

i.e. your second approach seems to be the right onw. However we should also do a quick sanity check in here and see that $\sum_{k=1}^\infty P[X=k] = 1$ however, I am too silly to prove that :-( All I got is

\begin{align*} \sum_{k=1}^\infty P[X=k] &= \sum_{k=1}^\infty (1-\theta^k)\theta^{k(k-1)/2} \\ &\leq \sum_{k=1}^\infty (1-0)\theta^{k(k-1)/2} = \sum_{k=1}^\infty \theta^{k(k-1)/2} \\ &\leq \sum_{k=1}^\infty \theta^{k(k-1)/2} + \sum \text{missing exponents} \\ &= \sum_{k=1}^\infty \theta^k = \frac{1}{1-\theta} \end{align*}

However, a little program that I wrote tells me that this sum converges absolutely quickly against 1 (after 10 summands already for theta=0.5)... so this seems to be the right answer I guess...

EDIT: ah... sometimes it is so simple...

\begin{align*} \sum_{k=1}^\infty (1-\theta^k)\theta^{k(k-1)/2} &= \sum_{k=1}^\infty \theta^{k(k-1)/2} - \sum_{k=1}^\infty \theta^{k + k(k-1)/2} \end{align*}

and surprisingly (as I did not see :-))

$$k + \frac{k(k-1)}{2} = \frac{k(k+1)}{2}$$

so the right hand side kills all the summands except for the first one which is 1... mystery solved, eh?