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Consider the following argument:

If $(X| Y \ \text{OR} \ Z)$ is true, $(X| Y)$ must be true.

For example, if $f(t)=10 $ when $ t=1 $ or $ \ t=0$ is true, then $f(t)=10 $ when $ t=1$ must be true as well.

Therefore, the set of $X$ that make $(X|Y,Z)$ true is a subset of the set of $X$ that make $(X|Y)$ true.

Therefore, $P(X|Y,Z) \le P(X|Y)$

Is this argument correct?

Glen_b
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Dr. Who
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  • Welcome Dr. Who. I have edited your question to display math as math, you should check that I have not introduced any errors. – Antoine Vernet Jul 22 '16 at 06:16

1 Answers1

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I assume that in your notation "P(X|Y)" is a conditional probability and that "<=" is an inequality and not an implication. Otherwise, please forget my answer, clarify the question and edit the tags.

I'll give a simple counterexample.

Let's suppose we take a random animal from a hen house with 50 chickens and 50 rabbits, and let's define the following events:

X=The animal has wings.

Y=The animal is a rabbit.

Z=The animal is a chicken.

P(X|Y) = P(X and Y)/P(Y) = (0/100)/(50/100) = 0/50 = 0

P(X|Y or Z) = P(X and (Y or Z))/P(Y or Z) = (50/100)/(100/100) = 50/100 = 0.5

Then P(X|Y or Z) = 0.5 > 0 = P(X|Y)

Pere
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  • Thanks for the counterexample! After some thoughts, I think the following altered statement may be true. P(X|Y OR Z) is always smaller than or equal to the max of P(X|Y) and P(X|Z). What do you think? – Dr. Who Jul 22 '16 at 15:12
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    I'm afraid it's false, too. Let's see the experiment of taking a random number from the set {1,2,3,4}. X={2,4}, Y={1,2}, Z={1,4}. P(X|Y)=P(X|Z)=1/2 but P(X|Y or Z)=2/3. – Pere Jul 22 '16 at 16:17
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    You are right. You win :) – Dr. Who Jul 22 '16 at 16:31