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I'm learning probability theory and I am quite new in the concept. I'm stuck with the following problem:

Consider a situation where people often get bitten by dogs (just as an example).

Let $p_A(n)$ be the probability that a person A is bitten on day $n$, given that he/she hasn't been bitten on day $1, ..., n-1$.

Let $X_A$ be the number of the day when person A is bitten for the first time.

If I know that $p_A(n) = \frac{1}{n+1}$, what is $\mathbb{P}(X_A=n)$?

I just can't really get my head around the meaning of the pmf for example. In my eyes it would look like $\mathbb{P}(X_A=n) = \frac{1}{n+1}$. Is this correct? If not, how am I interpreting it wrong and does anyone have a hint on what it should be?

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I don't like the notation for $p_A(n)$, here, because it hides the fact that this is a conditional probability.

The event $X_A=n$ is the event that they are not bitten on days $1,\ldots,n-1$, but are bitten on day $n$. On the other hand, $p_A(n)$ is the probability that they are bitten on day $n$ ASSUMING they were not bitten previously.

Let's work out a few to point out the difference. Let $B_i$ be the indicator for whether or not a bite occurred on day $i$. Note that this means we can rewrite $$ p_A(n)=P(B_n=1\mid B_1=0,B_2=0,\ldots,B_{n-1}=0) $$

Since day 1 is the first day, the two agree: $P(X_A=1)=\frac{1}{2}$.

Now, $X_A=2$ if and only if they are NOT bitten on day 1 and ARE bitten on day 2. We can write this as $$ P(X_A=2)=P(B_1=0, B_2=1). $$ The only way we can compute this is by leveraging our conditional probability and the previous result: $$ P(X_A=2)=P(B_1=0,B_2=1)=P(B_2=1\mid B_1=0)\cdot P(B_1=0)=p_A(2)\cdot\left(1-\frac{1}{2}\right)=\frac{1}{6}. $$

Similarly, for $P(X_A=3)$: $$ P(X_A=3)=P(B_1=B_2=0,B_3=1)=P(B_3=1\mid B_1=B_2=0)\cdot P(B_1=B_2=0). $$ The first term is precisely $p_A(3)=\frac{1}{4}$; the second is precisely the probability that $X_A\geq 3$, which you can figure out using the above.

Can you see how to keep going? What patterns emerge?

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$\mathbb{P}(X_A=n)$ is the probability that the person did not get bitten on days $1, \ldots,n-1$ but got bitten on day $n$.

Think of $n=1$. The person got bitten on day 1. This happens with probability $p_A(1) = 1/2$.

Think of $n=2$. The person did not get bitten on day 1 and got bitten on day 2. This happens with probability $\mathbb{P}(\text{bitten on day } 2 \text{ and not bitten on day }1) = \mathbb{P}(\text{bitten on day} 2 | \text{not bitten on day } 1)\mathbb{P}(\text{not bitten on day} 1)$ = $p_A(2) \times {1 \over 2} = 1/6$. The middle step followed from Baye's rule.

Continuing, you should be able to set up the recursion:

$$\mathbb{P}(X_A=n) = p_A(n)\prod_{j=1}^{n-1} \mathbb{P}(X_A=j)$$

Not sure if there is a closed form expression possible.

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Refer to the diagram (where "Bn" and "B'n" denote "bitten on day n" and "not bitten on day n", respectively): enter image description here

For $n=1$: $$p_A(1)=\frac1{1+1}=\frac12, \\ \mathbb P(X_A=1)=\mathbb P(B1)=p_A(1)=\frac12.$$ For $n=2$: $$p_A(2)=\frac1{2+1}=\frac13,\\ \mathbb P(X_A=2)=\mathbb P(B'1)\cdot \mathbb P(B2)=(\underbrace{1-p_A(1)}_{1/2})\cdot \underbrace{p_A(2)}_{1/3}=\frac1{2\cdot 3}$$ For $n=3$: $$p_A(3)=\frac1{3+1}=\frac14,\\ \mathbb P(X_A=3)=\mathbb P(B'1)\cdot \mathbb P(B'2)\cdot \mathbb P(B3)=(\underbrace{1-p_A(1)}_{1/2})\cdot (\underbrace{1-p_A(2)}_{2/3})\cdot \underbrace{p_A(3)}_{1/4}=\frac1{3\cdot 4}$$

Can you generalize it for any $n$?

Answer:

$\mathbb P(X_A=n)=\frac{1}{n(n+1)}.$

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Thanks everyone!! It makes total sense to me now. I was indeed a bit confused by the fact that $p_A(n)$ is a conditional probability.

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  • $\begingroup$ and the award goes to... you should accept a suitable answer by ticking and close this question. Good luck! $\endgroup$ – farruhota May 23 at 14:29
  • $\begingroup$ how do I close this question? $\endgroup$ – bladiebla May 23 at 16:11
  • $\begingroup$ I see you did accept Nick Peterson’s answer. Also, you can upvote the helpful answers. You can answer your own question and accept your own answer. Your current answer is actually a comment, which should be placed in the body of your question or under other answers. $\endgroup$ – farruhota May 23 at 18:22

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