Calculating Joint Probability - Get the Answer!

[Usagi]

http://img151.imageshack.us/img151/6002/variance.jpg

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

Thanks.

 

[CaptainBlack]

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

 

[Usagi]

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 $P(X_1, X_2) = P(X_1, X_5)$

 

[CaptainBlack]

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

 

[Usagi]

Thanks :)