क्यों

Question

मैं मानता हूँ कि

$$P(A|B) = P(A | B,C) * P(C) + P(A|B,\neg C) * P(\neg C)$$

सही है, जबकि

$$P(A|B) = P(A | B,C) + P(A|B,\neg C)$$

गलत है।

हालाँकि, मुझे बाद में एक के बारे में "अंतर्ज्ञान" मिला है, यानी आप दो मामलों (C या Not C) को विभाजित करके प्रायिकता P (A | B) पर विचार करें। यह अंतर्ज्ञान गलत क्यों है?

Comments

अपने समीकरणों का परीक्षण करने के लिए यहां एक सरल उदाहरण दिया गया है। टॉस दो स्वतंत्र, निष्पक्ष सिक्के। बता दें कि $A$ ऐसी घटना है जिसमें पहली बार सिर आता है, $B$ ऐसी घटना है कि दूसरा सिर पर आता है, और $C$ ऐसा घटना है कि दोनों प्रमुख बन जाते हैं। है या तो समीकरण आप ने लिखा सही है?

— ए। रेक्स

कुल संभावना के कानून जैसे: अगर आप सशर्त संभावनाओं की राशि के रूप में बिना शर्त संभावना व्यक्त करना चाहता हूँ, तुम कंडीशनिंग घटना से वजन चाहिए कहते हैं $P(A) = P(A|B)P(B) + P(A|\bar{B})P(\bar{B})$

— Adamo

Answer 1

एक आसान काउंटर उदाहरण के रूप में मान लीजिए,, कि संभावना $P(A)$ के $A$ है $1$ , का मान की परवाह किए बिना $C$ । फिर, यदि हम गलत समीकरण लेते हैं, तो हमें यह मिलता है:

$P(A | B) = P(A | B, C) + P(A | B, \neg C) = 1 + 1 = 2$

यह स्पष्ट रूप से सही नहीं हो सकता है, शायद से अधिक नहीं हो सकता है । यह अंतर्ज्ञान का निर्माण करने में मदद करता है कि आपको दो मामलों में से प्रत्येक के लिए एक वजन निर्धारित करना चाहिए कि यह मामला कितना संभव है , जिसके परिणामस्वरूप पहला (सही) समीकरण होता है। ।$1$

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यह आपको अपने पहले समीकरण के करीब लाता है, लेकिन वजन पूरी तरह से सही नहीं है। सही वजन के लिए ए। रेक्स की टिप्पणी देखें।

Comments

"पहले (सही) समीकरण" में वजन होना चाहिए और पी ( ¬ सी ) , या वे होना चाहिए पी ( सी | बी ) और पी ( ¬ सी | बी ) ? $P(C)$$P(\neg C)$$P(C\mid B)$$P(\neg C\mid B)$

— ए। रेक्स

@ A.Rex एक अच्छा बात है कि, पूर्ण शुद्धता के लिए मुझे लगता है कि यह होना चाहिए और पी ( ¬ सी | बी ) । समीकरण के बाईं ओर सब कुछ (सिर्फ एक शब्द) मान लेता है कि बी दिया गया है, इसलिए बिना किसी अतिरिक्त धारणा (जैसे कि बी और सी एक दूसरे से स्वतंत्र हैं), वही दाईं ओर होना चाहिए $P(C | B)$$P(\neg C | B)$$B$$B$$C$

— -हैंड

जरा सोचिए A | B का 200% होना निश्चित है।

— मार्क एल। स्टोन

@ MarkL.Stone इसका मतलब यह है कि यह हमेशा दो बार होता है? ;)

— मोनिका

Answer 2

डेनिस के जवाब में एक महान काउंटर-उदाहरण है, गलत समीकरण को बाधित करना। यह उत्तर यह बताना चाहता है कि निम्नलिखित समीकरण सही क्यों है:

$$P(A|B) = P(A|C,B) P(C|B) + P(A|\neg C,B) P(\neg C|B).$$

जैसा कि प्रत्येक शब्द पर वातानुकूलित है , हम B द्वारा संपूर्ण संभाव्यता स्थान को बदल सकते हैं और B पद छोड़ सकते हैं । यह हमें देता है:$B$$B$$B$

$$P(A) = P(A|C) P(C) + P(A|\neg C) P(\neg C).$$

तो फिर तुम क्यों इस समीकरण है पूछ रहे हैं और पी ( ¬ सी $P(C)$$P(\neg C)$ उस में शर्तें।

कारण यह है कि है का एक हिस्सा है एक में सी और पी ( एक | ¬ सी ) पी ( ¬ सी ) का एक हिस्सा है एक में ¬ सी और दो को जोड़ने के एक । आरेख देखें। दूसरी तरफ P ( A | C ) A और P ( A ) युक्त C का अनुपात है$P(A|C) P(C)$$A$$C$$P(A|\neg C) P(\neg C)$$A$$\neg C$$A$$P(A|C)$$C$$A$ का अनुपात है ¬ सी युक्त एक - तो वे आम हरों की जरूरत नहीं है इसलिए उन्हें जोड़ने व्यर्थ है इन अलग-अलग क्षेत्रों के अनुपात में कर रहे हैं।$P(A|\neg C)$$\neg C$$A$

Comments

Not "everything is conditioned on $B$". In particular, $P(C)$ and $P(\neg C)$ are not, so you can't just drop $B$. Moreover, this might suggest the equation is wrong!

— A. Rex

@A.Rex Technically you're right, I should have said every term involving $A$ is conditioned on $B$ (I made a simple substitution $A|B \rightarrow A$). I will correct the answer.

— Reinstate Monica

My objection wasn't a technicality. Your diagram correctly proves that $P(A) = P(A\mid C) P(C) + P(A\mid\neg C) P(\neg C)$, which after conditioning on $B$ becomes $P(A\mid B) = P(A\mid B,C) P(C\mid B) + P(A\mid B,\neg C) P(\neg C\mid B)$; note that the probabilities of $C$ and $\neg C$ are also conditioned on $B$. This is not the first equation given in the OP, which is good news, because the first equation given in the OP is not correct.

— A. Rex

@A.Rex You are right once again, $C$ must also conditioned on $B$ as the proportion of the probability space contained in $C$ might not be the same as the proportion of $B$ contained in $C$. This point escaped me. I will revise again.

— Reinstate Monica

Answer 3

I know you've already received two great answers to your question, but I just wanted to point out how you can turn the idea behind your intuition into the correct equation.

First, remember that $P(X \mid Y) = \dfrac{P(X \cap Y)}{P(Y)}$ and equivalently $P(X \cap Y) = P(X \mid Y)P(Y)$.

To avoid making mistakes, we will use the first equation in the previous paragraph to eliminate all conditional probabilities, then keep rewriting expressions involving intersections and unions of events, then use the second equation in the previous paragraph to re-introduce the conditionals at the end. Thus, we start with:

$$P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}$$

We will keep rewriting the right-hand side until we get the desired equation.

The casework in your intuition expands the event $A$ into $(A \cap C) \cup (A \cap \neg C)$, resulting in

$$P(A \mid B) = \dfrac{P(((A \cap C) \cup (A \cap \neg C)) \cap B)}{P(B)}$$

As with sets, the intersection distributes over the union:

$$P(A \mid B) = \dfrac{P((A \cap B \cap C) \cup (A \cap B \cap \neg C))}{P(B)}$$

Since the two events being unioned in the numerator are mutually exclusive (since $C$ and $\neg C$ cannot both happen), we can use the sum rule:

$$P(A \mid B) = \dfrac{P(A \cap B \cap C)}{P(B)} + \dfrac{P(A \cap B \cap \neg C)}{P(B)}$$

We now see that $P(A \mid B) = P(A \cap C \mid B) + P(A \cap \neg C \mid B)$; thus, you can use the sum rule on the event on the event of interest (the "left" side of the conditional bar) if you keep the given event (the "right" side) the same. This can be used as a general rule for other equality proofs as well.

We re-introduce the desired conditionals using the second equation in the second paragraph:

$$P(A \cap (B \cap C)) = P(A \mid B \cap C)P(B \cap C)$$ and similarly for $\neg C$.

We plug this into our equation for $P(A \mid B)$ as:

$$P(A \mid B) = \dfrac{P(A \mid B \cap C)P(B \cap C)}{P(B)} + \dfrac{P(A \mid B \cap \neg C)P(B \cap \neg C)}{P(B)}$$

Noting that $\dfrac{P(B \cap C)}{P(B)} = P(C \mid B)$ (and similarly for $\neg C$), we finally get

$$P(A \mid B) = P(A \mid B \cap C)P(C \mid B) + P(A \mid B \cap \neg C)P(\neg C \mid B)$$

Which is the correct equation (albeit with slightly different notation), including the fix A. Rex pointed out.

Note that $P(A \cap C \mid B)$ turned into $P(A \mid B \cap C)P(C \mid B)$. This mirrors the equation $P(A \cap C) = P(A \mid C)P(C)$ by adding the $B$ condition to not only $P(A \cap C)$ and $P(A \mid C)$, but also $P(C)$ as well. I think if you are to use familiar rules on conditioned probabilities, you need to add the condition to all probabilities in the rule. And if there's any doubt whether that idea works for a particular situation, you can always expand out the conditionals to check, as I did for this answer.

Comments

+1. I think you extracted the equation that OP tried to intuit: $P(A\mid B)=P(A\cap C\mid B)+P(A\cap\neg C\mid B)$.

— A. Rex

Thanks! That was the main point I wanted to make, but couldn't figure out a high-level explanation why the intersection goes on the left rather than the right, so I used formulas instead. Also, I just noticed you were the one who pointed out the mistake in OP's formula, so I credited you for that. (I probably wouldn't have noticed either, lol.)

— YawarRaza7349

Answer 4

Probabilities are ratios; the probability of A given B is how often A happens within the space of B. For instance, $P(\text{rain|March})$ is the number of rainy days in March divided by the number of total days in March. When dealing with fractions, it makes sense to split up numerators. For instance,

$$\begin{align} P(\text{rain or snow|March}) &= \frac{(\text{number of rainy or snowy days in March})}{(\text{total number of days in March})} \\[7pt] &= \frac{(\text{number of rainy days in March})}{(\text{total number of days in March)}} + \\[4pt] &\qquad \frac{\text{(number of snowy days in March)}}{(\text{total number of days in March)}} \\[7pt] &= P(\text{rain|March})+P(\text{snow|March}) \end{align}$$

This of course assumes that "snow" and "rain" are mutually exclusive. It does not, however, make sense to split up denominators. So if you have $P(\text{rain|February or March})$, that is equal to

$$\frac{(\text{number of rainy days in February and March})}{(\text{total number of days in February and March})}.$$

But that is not equal to

$$\frac{(\text{number of rainy days in February})}{(\text{total number of days in February})} + \frac{(\text{number of rainy days in March)}}{(\text{total number of days in March)}}.$$

If you're having trouble seeing that, you can try out some numbers. Suppose there are 10 rainy days in February and 8 in March. Then we have

$$\frac{(\text{number of rainy days in February and March})}{(\text{total number of days in February and March)}} = (10+8)/(28+31) = 29.5\%$$

and

$$\begin{align} \frac{(\text{number of rainy days in February})}{(\text{total number of days in February)}} + \frac{(\text{number of rainy days in March)}}{(\text{total number of days in March)}} &= (10/28)+(8/31) \\ &= 35.7\% + 25.8\% \\ &= 61.5\% \end{align}$$

The first number, 29.5%, is the average of 35.7% and 25.8% (with the second number weighted slightly more because there is are more days in March). When you say $P(A|B)=P(A|B,C)+P(A|B,¬C)$ you're saying that $\frac{x_1+x_2}{y_1+y_2} = \frac{x_1}{y_1}+\frac{x_2}{y_2}$, which is false.

Answer 5

If I go to Spain, I can get sunburnt.

$$P(sunburnt | Spain)=0.2$$ This tells me nothing about getting sunburnt if not going to Spain, let's say $$P(sunburnt|\neg Spain)=0.1$$ This year I'm going to Spain, so $$P(sunburnt)=0.2$$ Letting $B=\Omega$, this is, $P(B)=1$, your intuition would imply $$P(A)=P(A|C)+P(A|\neg C)$$ which by the previous argument, isn't neccesarily true.