this is some lecture slides from my school. I don't understand why we need to sum up all from $m=0$ to $m=+ \infty$. I think $m$ is a fixed number which represents the number of type $2$ event?
plz help me! many thanks in advance!
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Franciele Daltoé
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whoisit
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Recall the rule of average conditional probabilities:
For a partition $B_1,\dotsc,B_n$ of an outcome space $\Omega$, $$P(A) = P(A|B_1)+\dotsb+P(A|B_n)P(B_n).$$ In your case \begin{align*} P(X_1 = n) &= P(X_1 = n|X = n+0)P(X = n+0)\\ &\qquad\qquad+P(X_1 = n|X = n+1)P(X = n+1)\\ &\qquad\qquad+P(X_1 = n|X = n+2)P(X = n+2)+\dotsb\\ &= \sum_{m=0}^\infty P(X_1 = n|X=n+m)P(X = n+m). \end{align*} Then, just plug in the given information.
Em.
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I don't understand why we need to use average conditional probabilities. For example, there are 10 customers(4 males and 6 females). If I want to work out the probability for p(males=4|total=10), why do I need to start with female customer from 0,1,...to 6, there are just 6 females which is the condition. I know I must have some fundamentally mistakes here, it's driving insane. – whoisit Dec 03 '15 at 09:59
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That is not what the result is trying to indicate. It is trying to show that the unconditional distribution of $X_1$ (males) follows a $\text{Pois}(.5\cdot 12).$ – Em. Dec 03 '15 at 10:11
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Using your example, we are looking for $P(X_1) = 4$, not what you wrote. The point is that when looking for $P(X = 4)$, you don't know how many women there were. That is why you have to condition on all possible number of women that could have arrived. Thus, it is incorrect to say that there were 6 women. When looking for $P(X_1) = 4$, you don't know how many woman there were. – Em. Dec 03 '15 at 10:34
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Sure, vote up\check would be nice. Also notice in the example given in the picture at the bottom, the shop owner expects 12 customers per hour. It's not saying that there were exactly 12 customers. – Em. Dec 03 '15 at 10:48
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it;s a little bit clearer to me. But I'm still wondering, if it's an average probability, why don't we go like P(X1)/m, it's more like a cumulative probability rather than average probability. And yes we don't know how many women arrived when we P(X1)=4, but the number of women should only be a fixed number, if we don't know how much is it, we can just use a variable m? – whoisit Dec 03 '15 at 10:48
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It is not an average probability. Also the number of women that arrive in the next hour to the shop is not fixed. It could be 0, or 1, or 2, or 3, or 4, or .... That is why you sum up every case. – Em. Dec 03 '15 at 10:56
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It is essential that:$$\{X_1=n\}=\bigcup_{m=1}^{\infty}\{X_1=n\wedge X=m\}$$ Note that the events on RHS are disjoint. So this gives: $$P(X_1=n)=\sum_{m=0}^{\infty}P(X_1=n\wedge X=m)$$ Here: $$P(X_1=n\wedge X=m)=P(X_1=n\mid X=m)P(X=m)$$
For $m<n$ the events are empty so you can start with $m=n$.
Finally: $$P(X_1=n)=\sum_{m=n}^{\infty}P(X_1=n\mid X=m)P(X=m)=$$$$\sum_{m=0}^{\infty}P(X_1=n\mid X=n+m)P(X=n+m)$$
drhab
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I feel I need an example to absorb it. For example, there are 10 customers(4 males and 6 females). If I want to work out the probability for p(males=4|total=10), why do I need to start with female customer from 0,1,...to 6, there are just 6 females which is the condition. – whoisit Dec 03 '15 at 10:01
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p(males=4|total=10)=$\binom{10}4p^4(1-p)^6$ (binomial distribution with $n=10$ and $p$ is the probabiltity that the customer is male). No need for looking at $0,1,2,\dots$ female customers at all. That only comes in if you are looking for p(males=4). By the calculation we split up in males=4 and females=0, males =4 and females =1,... et cetera. – drhab Dec 03 '15 at 12:57
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In this case, we are trying to show
X ~ Poisson(lambda=12)
p = 0.5
X1 ~ Poisson(lambda=12*0.5=6)
Let's say we want to find P(X1=4). There are many ways to get that
P(X1=4|X=0) = 0
P(X1=4|X=1) = 0
P(X1=4|X=2) = 0
P(X1=4|X=3) = 0
P(X1=4|X=4) = $\binom{4}{4} 0.5^4 \cdot 0.5^0$
P(X1=4|X=5) = $\binom{5}{4} 0.5^4 \cdot 0.5^1$
P(X1=4|X=6) = $\binom{6}{4} 0.5^4 \cdot 0.5^2$
P(X1=4|X=j) = $\binom{j}{4} 0.5^4 \cdot 0.5^{j-4}$ for $j>=4$
Applying the Law of Total Probability (https://en.wikipedia.org/wiki/Law_of_total_probability),
P(X1=4) = $\sum_{j=4}^{\infty} \binom{j}{4}0.5^4 \cdot 0.5^{j-4} \cdot P(X=j)$
Hopefully this explains why the summation is necessary.
