Conditional Probability/Restrictions

• Oct 20th 2013, 09:36 AM
ishihara
Conditional Probability/Restrictions
I'm looking to get some insight into a moderately challenging conditional probability problem:

Consider a sequence of random variables $\displaystyle X_1$,$\displaystyle X_2$...,$\displaystyle X_n$ which each take the values 0 and 1. Assume that
Pr($\displaystyle X_j$ = 1) = 1 - Pr($\displaystyle X_{j-1}$= 0) = $\displaystyle \phi$, $\displaystyle \hspace{5mm}$j =1,...,n

where 0 < $\displaystyle \phi$ < 1 and that

$\displaystyle \hspace{10mm}$Pr($\displaystyle X_j$ = 1|$\displaystyle X_{j-1}$=1) = $\displaystyle \lambda$, $\displaystyle \hspace{5mm}$j = 2,...,n.

(a) Find Pr($\displaystyle X_j$= 0|$\displaystyle X_{j-1}$=1), Pr($\displaystyle X_j$= 1|$\displaystyle X_{j-1}$=0),Pr($\displaystyle X_j$ = 0|$\displaystyle X_{j-1}$=0).
(b) Find the requirements on $\displaystyle \lambda$ so that this describes a valid probability distribution for $\displaystyle X_1$,$\displaystyle X_2$...,$\displaystyle X_n$.

So for part (a) I have: 1-$\displaystyle \lambda$, $\displaystyle \phi$, and 1-$\displaystyle \phi$, respectively.
I do not know how to begin part (b).
• Oct 20th 2013, 10:27 AM
SlipEternal
Re: Conditional Probability/Restrictions
For part (a), you may have some mistakes. Here is what I get:

$\displaystyle \mbox{Pr}(X_{j-1}=1) = \phi$

$\displaystyle \mbox{Pr}(X_{j-1}=0) = 1-\mbox{Pr}(X_{j-1}=1) = 1-\phi$

$\displaystyle \mbox{Pr}(X_j=1|X_{j-1}=1) = \lambda = \dfrac{\mbox{Pr}((X_j=1)\cap (X_{j-1}=1))}{\mbox{Pr}(X_{j-1}=1)} = \dfrac{\mbox{Pr}((X_j=1)\cap (X_{j-1}=1))}{\phi}$ implies

$\displaystyle \mbox{Pr}((X_j=1)\cap (X_{j-1}=1)) = \lambda\phi$

$\displaystyle \mbox{Pr}(X_j=0|X_{j-1}=1) + \mbox{Pr}(X_j=1|X_{j-1}=1) = 1$ so I agree that $\displaystyle \mbox{Pr}(X_j=0|X_{j-1}=1) = 1-\lambda$.

But, $\displaystyle \mbox{Pr}(X_j=1|X_{j-1}=0) = \dfrac{\mbox{Pr}((X_j=1)\cap (X_{j-1}=0))}{\mbox{Pr}(X_{j-1}=0)} = \dfrac{\mbox{Pr}((X_j=1)\cap (X_{j-1}=0))}{1-\phi}$

Since $\displaystyle \phi = \mbox{Pr}(X_j=1) = \mbox{Pr}((X_j=1)\cap (X_{j-1}=0)) + \mbox{Pr}((X_j=1)\cap (X_{j-1}=1))$ $\displaystyle = \mbox{Pr}((X_j=1)\cap (X_{j-1}=0)) + \lambda\phi$, we find $\displaystyle \mbox{Pr}((X_j=1)\cap (X_{j-1}=0)) = \phi(1-\lambda)$

So, $\displaystyle \mbox{Pr}(X_j=1|X_{j-1}=0) = \dfrac{\mbox{Pr}((X_j=1)\cap (X_{j-1}=0))}{1-\phi} = \dfrac{\phi(1-\lambda)}{1-\phi}$

Then $\displaystyle \mbox{Pr}(X_j=0|X_{j-1}=0) = 1- \mbox{Pr}(X_j=1|X_{j-1}=0) = 1-\dfrac{\phi(1-\lambda)}{1-\phi} =$ $\displaystyle \dfrac{1-2\phi + \phi\lambda}{1-\phi}$

Let's check that we have everything correct:

So, based on this, we have

\displaystyle \begin{align*}\mbox{Pr}((X_j=1)\cap (X_{j-1}=0)) & = \phi(1-\lambda) \\ \mbox{Pr}((X_j=1)\cap (X_{j-1}=1)) & = \phi\lambda \\ \mbox{Pr}((X_j=0)\cap (X_{j-1}=0)) & = 1-2\phi + \phi\lambda \\ \mbox{Pr}((X_j=0)\cap (X_{j-1}=1)) & = \phi(1-\lambda)\end{align*}
• Oct 20th 2013, 10:49 AM
ishihara
Re: Conditional Probability/Restrictions
Thanks for the help.

Did you have any idea on the restrictions for $\displaystyle \lambda$?

Obviously the entire set has to be equal to a probability of 1. So are the bounds for $\displaystyle \lambda$ are 0 < $\displaystyle \lambda$ < 1?
• Oct 20th 2013, 11:20 AM
SlipEternal
Re: Conditional Probability/Restrictions
Quote:

Originally Posted by ishihara
Thanks for the help.

Did you have any idea on the restrictions for $\displaystyle \lambda$?

Obviously the entire set has to be equal to a probability of 1. So are the bounds for $\displaystyle \lambda$ 0 < $\displaystyle \lambda$ < 1?

Let's check out all of the restrictions:

From $\displaystyle \mbox{Pr}((X_j=1)\cap (X_{j-1}=0)) = \phi(1-\lambda)$:

$\displaystyle 0 \le \phi(1-\lambda) \le 1 \Rightarrow 0 \le 1-\lambda \le \dfrac{1}{\phi} \Rightarrow 1-\dfrac{1}{\phi} < 0 \le \lambda \le 1$

From $\displaystyle \mbox{Pr}((X_j=1)\cap (X_{j-1}=1)) = \phi\lambda$:

$\displaystyle 0 \le \lambda\phi \le 1 \Rightarrow 0 \le \lambda \le 1 < \dfrac{1}{\phi}$

From $\displaystyle \mbox{Pr}((X_j=0)\cap (X_{j-1}=0)) = 1-2\phi + \phi\lambda$:

$\displaystyle 0 \le 1-2\phi + \phi\lambda \le 1 \Rightarrow 2\phi-1 \le \phi\lambda \le 2\phi \Rightarrow 2-\dfrac{1}{\phi} \le \lambda \le 1 < 2$

This inequality gives a possible bound for $\displaystyle \lambda$. If $\displaystyle \dfrac{1}{2} < \phi < 1$, then $\displaystyle 0 < 2-\dfrac{1}{\phi} < 1$.

The conditional probabilities all give the same bounds for $\displaystyle \lambda$, as well.

I don't know if there are any other restrictions to consider.