# Proof of total probability theorem

• Dec 22nd 2009, 12:37 PM
rpatel
Proof of total probability theorem
Hi

My question is

State and prove the theorem of total probability ?

I tried looking for this proof in my textbook but i can't seem to find it anywhere

any help is appreichated

thanks
• Dec 22nd 2009, 12:56 PM
firebio
Quote:

Originally Posted by rpatel
Hi

My question is

State and prove the theorem of total probability ?

I tried looking for this proof in my textbook but i can't seem to find it anywhere

any help is appreichated

thanks

Are you looking for the proof of this theorem? i wasnt sure

http://mathworld.wolfram.com/images/...dEquation1.gif

I am not sure if i can post external links on this website...

Check out
Baye's Theorem explained with proof | TutorVista

it has a proof of this theorem
• Dec 22nd 2009, 02:39 PM
rpatel
are bayes theorem and theorem of total probability the same thing ?

I though total probability theorem had something to do with 'paritions'.
• Dec 22nd 2009, 02:44 PM
firebio
Quote:

Originally Posted by rpatel
are bayes theorem and theorem of total probability the same thing ?

I though total probability theorem had something to do with 'paritions'.

They are different things

in the website scroll down to Law of total probability
• Dec 23rd 2009, 12:25 AM
Moo
Hello,

Law of total probability :
For a partition $\displaystyle (A_n)_{n\geq 1}$ of $\displaystyle \Omega$, the probability space, for any event B (and for higher levels, considering the measured space $\displaystyle (\Omega,\mathcal{A},\mathbb{P})$, for any $\displaystyle B\in\mathcal{A}$...), we have :

$\displaystyle \mathbb{P}(B)=\sum_{n=1}^\infty \mathbb{P}(B|A_n)\mathbb{P}(A_n)$

and the proof.... :

$\displaystyle \bigcup_{n=1}^\infty A_n=\Omega$

So $\displaystyle B=B\cap \Omega=B\cap \bigcup_{n=1}^\infty A_n=\bigcup_{n=1}^\infty \{B\cap A_n\}$

But the last union is a disjoint union : $\displaystyle \forall i\neq j~,~ \{B\cap A_i\}\cap \{B\cap A_j\}=\emptyset$ (very easy to prove)

Thus the probability of the union is just the sum of the probabilities :

$\displaystyle \mathbb{P}(B)=\sum_{n=1}^\infty \mathbb{P}(B\cap A_n)$

Then just use the formula $\displaystyle \mathbb{P}(A\cap B)=\mathbb{P}(B|A)\mathbb{P}(A)$