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Problem: a person wants to use ATM. With equal probability either $0$ or $1$ person is ahead. If there is $1$ person, the waiting time is distributed exponentially with the $\lambda$. What's the CDF of the waiting time?

Solution (by the book): I do understand the general approach:

$T$ - the waiting time, $F_{T}(t)=P\{T\leq t\} = ?$

$C$ - the number of people ahead ($0$ or $1$)

$P\{T\leq t\}=P(C=0)*P\{T\leq t|C=0\}+P(C=1)*P\{T\leq t|C=1\}$

The book says that $P\{T\leq t|C=0\}=1$. Intuitively I do understand it: If no one's using the ATM, the probability that the waiting time is less or equal to a non-negative number is $0$. However, for the sake of understanding the formula I'd like to apply the definition of conditional probability here and compute $P\{T\leq t|C=0\}$ using the general fact that $P(A|B)=P(A\cap B)/P(B)$

How to compute the $P(\{T\leq t \cap C=0\})?$ Same question for the second part: How to compute the $P(\{T\leq t \cap C=1\})?$

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    $\begingroup$ Don't you have to have $P(T \leq t, C = 0)=1/2$, since $P(C=0)=1/2$? $\endgroup$
    – aduh
    May 7, 2017 at 21:06
  • $\begingroup$ @aduh I do, but here I'd like to know the reasoning that comes not from $1/2 : 1/2 = 1$, but from the concepts and their definition. Suppose I wasn't told that $P\{T\leq t|C=0\}=1$. How could I reason here? $\endgroup$ May 7, 2017 at 21:10

2 Answers 2

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Using the definition of conditional probability

$$P(T\le t \mid C=0)=\frac{P(T\le t \cap C=0)}{P(C=0)}=\frac{P(T\le t\cap C=0)}{P(C=0)}.$$

Furthermore

$$P(T\le t\cap C=0)=P(T\le t\cap C=0\cap T=0)=$$ $$=\begin{cases}P(C=0)&\text{ if } &t\ge 0\\0&\text{ otherwise}\end{cases}$$

because $\{C=0\}\subset\{T=0\}$ and, as a result $\{C=0\}\cap\{T=0\}=\{C=0\}$ and $\{T\le t\}\cap \{T=0\}\cap\{C=0\}=\{C=0\}$ if $t\ge0$ and it is $\emptyset$ otherwise.

But I would rather simply say that

$$P(T\le t\mid C=0)=\begin{cases}1&\text{ if }& t\ge 0\\0&\text{ if }& t<0\end{cases}$$

since I do know that if $C=0$ is given then the waiting time is $0$.

Also, $$P(T\le t\mid C=1)=\begin{cases}1-e^{\lambda t}&\text{ if }& t\ge 0\\0&\text{ if }& t<0.\end{cases}$$

So,

$$F_T(t)=\begin{cases}P(C=0)+(1-e^{\lambda t})P(C=1)&\text{ if }& t\ge 0\\0&\text{ if }& t<0.\end{cases}$$ For $\lambda=1$, $P(C=0)=\frac12=P(C=1)$ we have

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  • $\begingroup$ Why assume independence? $\endgroup$
    – aduh
    May 7, 2017 at 21:39
  • $\begingroup$ @adu: You are right we don't have to assume independence. I corrected my answer. $\endgroup$
    – zoli
    May 7, 2017 at 22:18
  • $\begingroup$ @zoli Thank you for the detailed explanation! But can't I prove the same way that $P(T≤t∣C=1)=1$ using the fact that $\{C=1\}\subset\{T=t\}$ and $\{C=1\}\cap\{T=t\}=\{C=1\}$, which will also give me $1$ when I divide by $P\{C=1\}$? $\endgroup$ May 8, 2017 at 7:41
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I suppose you could reason as follows.

Fix $t \geq 0$. Clearly, $\{T \leq t \cap C=0 \} \subseteq \{C=0 \}$. Conversely, if $C=0$, then $T=0$, which implies $T \leq t$. Therefore, the events $\{T \leq t \cap C=0 \}$ and $\{C=0 \}$ are identical and must have the same probability.

Though I think the reasoning with conditional probabilities is more natural.

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