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Multivariate Normal conditional tail Expextation

  1. Sep 15, 2012 #1
    Hi all,
    I need help regarding the following expression:
    x1 is a one dimension normal rv
    X is multivariate normal rv with n components: x1, x2,..., x_n
    K is a n dimension constant vector with n components: k1,k2,...,k_n
    X>K <==> x1>k1,x2>k2,...,x_n>k_n

    I know there is a closed form for the Bi-variate case. Is there one for this too?

    Thanks a lot!

  2. jcsd
  3. Sep 15, 2012 #2


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    Hey alondo and welcome to the forums.

    Are your variables independent or do they have some kind other relationship that involves a generic covariance matrix?
  4. Sep 15, 2012 #3
    Hi and thanks for your reply,
    The variables are dependent, and the covariance matrix can be calculated.
    However, I am not sure how to deal with the multiple conditions of the expectation:

    Thanks, and hope you can help.

  5. Sep 15, 2012 #4


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    In your multiple integral, what kinds of limits were you thinking about using?

    The expectation is simply going to be x1*P(x1 = x|X>k) (with the right limits).

    If you expand the conditional probability, it means you are going to get something that is complex, since you have dependencies between your variables.

    We know that P(x1 = x and x1 > k) = P(x1 > k) (Since x1 > k is a proper subset of the event corresponding to x1).

    With that said, what you are actually doing is censoring part of the variable, since you are considering a subset of the entire state-space as opposed to the entire state-space.

    If you are finding the expectation of this subset, you need to treat this as one proper random variable with the Kolmogorov axioms holding for this random variable and that means that you will have to normalize this new subset random variable.

    If your variables though are dependent on each other, you will need to calculate the right limits corresponding to your inequalities and if your limits are basically "entangled" (which they will be for dependency based situations for multivariate joint distributions), then you will need to use some really nice algorithms to do this.

    But if you can do all this, then find a random variable that has a proper PDF and then take the expectation of that new variable.
  6. Sep 16, 2012 #5
  7. Sep 16, 2012 #6


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    The example of E[X1|X2 > z] is a lot different than what you have.

    You wrote E[X1|X > k] by the X includes X1. Did you mean this or did you make a mistake?

    The reason is that if this was not a mistake, then it means you are conditioning on something with respect to a subset on X1 and not X1 itself, which means you will need to find a new random variable and take the expectation of that.

    If you want to find something like E[X1|X > K] where X does not include X1 and you have the joint distribution for X1,X2,..,XN, then this is simply a matter of finding the limits for X2 up to XN by solving the region of integration given the structure of the limits and then plugging those in to a computer program which will do a numerical multi-variable integral calculation and give you a number that corresponds to the expectation.

    The routine to calculate the expectation is easy once you have the limits, but getting the limits require that you solve the region in the same way that you solve regions for double integrals when you have entangled limits, except that due to there being multiple variables, you can't rely on the geometric intuition that you had for double integrals so you need to use an algebraic approach, or (which I recommend) an algorithm that someone has coded up.
  8. Sep 16, 2012 #7
    It was not a mistake, but the bi-variate case: E[X1|X2 > z] still holds when X2 is correlated to X1.
  9. Sep 16, 2012 #8


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    In that case you need to first find a random variable that corresponds to your constraints where X1 > k1 so that you have a PDF where X1 is restricted to that region where this new subset has a PDF that fits the axioms.

    If you don't do this, your computations and calculations will be wrong and won't make sense.

    So figure out the subset and the PDF corresponding to X1 > k1 first so that you get a new variable X1* corresponding to the new X1 and then take the conditional expectation E[X1*|(X2* > k2, X3* > k3, ... Xn* > kn)].

    It does not make sense mathematically or intuitively to do an expectation on a probability space where the integral or summation of that space does not equal 1 and if you don't do the above, you'll get something that just doesn't make sense.
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