MHB Calculating Joint Probability - Get the Answer!

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The discussion revolves around understanding the calculation of joint probabilities, specifically how to derive the boxed expression in the provided image. Participants clarify that the joint probability of different variables can be shown to be equal due to the arbitrary nature of their labeling. It is noted that \(P(X_j=1,X_k=1)\) can be expressed in terms of \(P(X_1=1,X_2=1)\) and that this equality holds across different indices. The conversation emphasizes that the probabilities remain consistent regardless of the variable labels, allowing for permutations in the indices. Overall, the thread provides insights into the principles of joint probability calculations.
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http://img151.imageshack.us/img151/6002/variance.jpg

Just wondering, how did they get the part boxed in blue?

Thanks.
 
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Usagi said:
http://img151.imageshack.us/img151/6002/variance.jpg

Just wondering, how did they get the part boxed in blue?

Thanks.
The part in the box comes from the observation that:

\[P(X_j=1,X_k=1)=P(X_1=1,X_2=1), \ \ j\ne k\],

and that:

\[P(X_1=1,X_2=1)=P(X_1=1)+P(X_2=1|X_1=1)\]

CB
 
Thanks for that, I realized it too however how does one show that $P(X_j, X_i) = P(X_1, X_2) = P(X_1, X_5)$ and so on?

It isn't so obvious to me that $P(X_1, X_2) = P(X_1, X_5)$
 
Usagi said:
Thanks for that, I realized it too however how does one show that $P(X_j, X_i) = P(X_1, X_2) = P(X_1, X_5)$ and so on?

It isn't so obvious to me that $P(X_1, X_2) = P(X_1, X_5)$

The informal explanation is that as the labelling is essentialy arbitary the probabilities are all the same.

More formally:

\[P(X_1=1,X_2=1)=P(X_2=1)+P(X_2)P(X_1=1|P(X_2=1)\]

But \(P(X_2=1|X_1=1)=P(X_2=1|X_5=1)\) and \(P(X_2=1)=P(X_5=1)\). which look better but is still just an observation that the idexing is arbitary and you can just permute indices.

CB
 
Thanks :)
 
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