# Joint probability

#### Usagi

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

Thanks.

#### CaptainBlack

##### Well-known member 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

#### Usagi

##### Member
Thanks for that, I realised 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

##### Well-known member
Thanks for that, I realised 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

#### Usagi

##### Member
Thanks 