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Thread: Maximum of a set of independent normal random variables

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    Maximum of a set of independent normal random variables




    Consider a set of random variables X_1, X_2, X_3, \ldots, X_n with unit varience but different means: \mu_1, \mu_2, \mu_3, \ldots, \mu_n. What is the probability that X_i will have the maximum value of the set.

    My initial guess was:
    P[X_i > X_{i'}|_{\forall i\neq i'}] = P[X_i>X_1]\cdot P[X_i>X_2]\cdots

    But if one considers a simple case where there are 3 random variables with mean 0 then obviously the probability of one of them being the maximum is 1/3.
    But using the formula above will give a probablility of 0.5*0.5=0.25.
    P[X_1 > X_2] = 0.5 when \mu_1=0, \mu_2=0

    What am I doing wrong here? In the simple 3 variable example I suspect that the probabilities aren't independent or something, but I can't wrap my head around it.
    Anyone have any thoughts?

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    Here is the thing I be thinkin


    pr( X1 > X2 AND X1 > X3 ) = pr( X1 > X2 > X3 ONION X1 > X2 > X3)

    Is the sum of the two disjoint events;

    This accounts for two of the 3! equally likely outcomes of the eperiment

    1/3

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    Quote Originally Posted by fed1 View Post
    pr( X1 > X2 AND X1 > X3 ) = pr( X1 > X2 > X3 ONION X1 > X2 > X3)
    Onion? What's that mean?

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    Is it math operator or vegetable ?

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    LOL. Its a vegetable. Mightymergz, I'm a little confused about how you set up the problem though. If the distribution of each X has a different mean, wouldn't the probability of selecting the same number in each X be different?

    Going back to your example, if you have 3 random variables, and lets say that their means are 0,1,2, then the probability of one of them being the max would not be 1/3.

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    They were working on the simplified example where each has a mean of 0 to test his result against intuition.

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    Indeed, it is a simplified version of the main problem. Also I've never heard of the ONION operator.. Onions are delicious to be sure, but they are less useful for calculating probabilities?

    In the simplified example all events are equally proabable so there are only 3!=6 possible outcomes.
    X_1>X_2>X_3,
    X_1>X_3>X_2,
    X_2>X_1>X_3,
    X_2>X_3>X_1,
    X_3>X_2>X_1,
    X_3>X_1>X_2

    P[X1>X2,X3] is two out of the 6 possibilities so the probability is 1/3. Unfortunately this method doesn't work if the random variables have different means. Is there another method or can I modify the above method to work when all choices aren't equally probable..

    thanks for your help so far!

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    Can I ask what this is for? Just interested? Is it homework? Required for research?

    Do we know anything else about the variables? What distribution they have maybe?

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    Hey Dason, ya, this problem is related to some research I am involved with for the summer. I am an undergrad co-op student on a summer research position

    The problem I posted is just one way to see the problem, I suppose. The main problem is: consider a segment of a discrete signal from t=-T to T. The signal f(t) = A(t) + y(t) where A(t) is some known function and y(t) is a normal zero mean unit varience random variable.

    The way I posed the problem is to just take A(t) as the mean of the random variables at each point and then calculate the statistics from there. We want to know with what probability will each point in the time series be the maximum. P[f(t)>f(t')|_{\forall t'\neq t}]

    Some things I would expect would be that the point with the highest probability of being the maximum would be the point with the highest mean/max(A(t)) (duh)
    and normalization: ie the peak is somewhere:
    \sum_{t=-T}^T P[f(t)>f(t')|_{\forall t'\neq t}]\approx 1

    But the initial method I tried in the OP definitely doesn't satisfy the normalization requirement. I suppose I could try just adding a normalization constant, but that doesn't seem right at all..

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    Ok...here's my reasoning...though I'd love for someone to chime in:
    Given \; \mu _{1}<\mu _{2}<\mu _{3} \; and \; \sigma _{1}=\sigma _{2}=\sigma _{3}=\sigma
    P(X_{3}>X_{2},X_{1})=P(X_{2}<x_{3})*P(X_{1}<x_{3})=P(Z<\frac{x_{3}-\mu_{2}}{\sigma})*P(Z<\frac{x_{3}-\mu_{1}}{\sigma})

    So, if the mu's were 0,1,2 and the sigma was 1, then we'd have
    P(X_{3}>X_{2},X_{1}) = P(Z<\frac{x_{3}-1}{1})*P(Z<\frac{x_{3}-0}{1})

    If \; x_{3}=\mu_{3}=2, \; then \; P(X_{3}>X_{2},X_{1}
=P(Z<1)*P(Z<2)
    \approx 0.841345*0.977250=0.8222

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    Hi Link, Thanks for your thoughts. I think I follow your train of thought but I numerically calculated the probabilities for \mu's equal to 0,1,2 and the probability for X_3 to be the maximum is closer to 0.73ish

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    really? And that's with a sigma of 1? I'm having a problem seeing where I might have went wrong. Though I relied on an online calculator to get me the p-values, I don't see them noticeably wrong.

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    I think I've got an answer. It's not necessarily pretty but I think I can do some work to simplify it. It matches intuition on that if all have the same mean and if there are 3 r.v.s then the probability is 1/3, if there are n then it's 1/n. But like I said when the means aren't the same it's not pretty so I'm going to see if it simplifies at all.

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    Quote Originally Posted by mightymergz View Post
    Hi Link, Thanks for your thoughts. I think I follow your train of thought but I numerically calculated the probabilities for \mu's equal to 0,1,2 and the probability for X_3 to be the maximum is closer to 0.73ish
    I did some simulations and I the probability around .728 so I think something went wrong Link. I think that P(X1 > X2) and P(X1 > X3) aren't necessarily independent because if you know that X1 > X2 then it's probably more likely that X1 > X3. But that's just a thought.

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    @Link
    ya with var/stddev of 1. Someone suggested to me that P(X_{3}>X_{2},X_{1})=P(X_1<x_3)*P(X_2<x_3) aren't independent events. I.e. the probability of the second event depends on the result of the first one. i.e. the probability should be more like:
    P[X_3>X_2,X_1]=P[X_2<X_3]*P[X_1<X_3 | (X_2<X_3)]
    though I'm not sure how to calculate that second conditional probability.. In the case of the simplified example I know it should be 2/3 simply because the P[X_2<X_3]=0.5 and 1/2 * 2/3 = 1/3 which is what we expect

    @Dason
    Ya I think it is something like that

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