Results 1 to 4 of 4

Math Help - Conditional Probability/Restrictions

  1. #1
    Newbie
    Joined
    Sep 2013
    From
    U.S.A.
    Posts
    8

    Conditional Probability/Restrictions

    I'm looking to get some insight into a moderately challenging conditional probability problem:


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


    where 0 < \phi < 1 and that


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


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




    So for part (a) I have: 1- \lambda, \phi, and 1- \phi, respectively.
    I do not know how to begin part (b).
    Last edited by ishihara; October 20th 2013 at 10:32 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,880
    Thanks
    742

    Re: Conditional Probability/Restrictions

    For part (a), you may have some mistakes. Here is what I get:

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

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

    \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

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

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

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

    So, \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 \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} = \dfrac{1-2\phi + \phi\lambda}{1-\phi}

    Let's check that we have everything correct:

    So, based on this, we have

    \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*}
    Last edited by SlipEternal; October 20th 2013 at 10:51 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2013
    From
    U.S.A.
    Posts
    8

    Re: Conditional Probability/Restrictions

    Thanks for the help.

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

    Obviously the entire set has to be equal to a probability of 1. So are the bounds for \lambda are 0 < \lambda < 1?
    Last edited by ishihara; October 20th 2013 at 10:58 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,880
    Thanks
    742

    Re: Conditional Probability/Restrictions

    Quote Originally Posted by ishihara View Post
    Thanks for the help.

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

    Obviously the entire set has to be equal to a probability of 1. So are the bounds for \lambda 0 < \lambda < 1?
    Let's check out all of the restrictions:

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

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

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

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

    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 \lambda. If \dfrac{1}{2} < \phi < 1, then 0 < 2-\dfrac{1}{\phi} < 1.

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

    I don't know if there are any other restrictions to consider.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: July 22nd 2011, 01:39 AM
  2. Conditional Probability using the Law of Total Probability
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: October 7th 2010, 03:01 AM
  3. Conditional Probability
    Posted in the Statistics Forum
    Replies: 3
    Last Post: December 16th 2009, 06:27 PM
  4. Continuous probability - conditional probability
    Posted in the Statistics Forum
    Replies: 1
    Last Post: October 1st 2009, 01:21 AM
  5. Conditional Probability
    Posted in the Statistics Forum
    Replies: 1
    Last Post: January 25th 2008, 02:22 PM

Search Tags


/mathhelpforum @mathhelpforum