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I Can these equations be solved analytically?

  1. Jun 26, 2016 #1
    Hi everyone,
    I am studying a problem related to Bayesian probability, and I came across two equations, which as far as I can tell can only be solved numerically, but as I'm no expert I would like to hear your opinion, please.

    The first one is:

    [itex]P(a) \cdot \left[ 1 - \Phi \left( \frac {x - \mu_a} { \sqrt 2 \cdot \sigma_a} \right) \right] = (1 - P(a) ) \cdot \Phi \left( \frac {x-\mu_b} { \sqrt 2 \cdot \sigma_b} \right) [/itex]


    [itex] \Phi (y) = \frac {1} {\sqrt {2 \pi}} \cdot \int_{- \inf}^y {e^{-t^2 / 2}} \, dt = \frac 1 2 \cdot \left[ 1 + {erf} \left( \frac {y} {\sqrt 2} \right) \right] [/itex]

    P(a) is a probability, thus a real (?) number between 0 and 1, and I need to solve for x.

    Initially I had no doubt that this could not be solved analytically for x. But then as I was reading something about statistical power, in an example they showed how you can invert Φ using a 'probit' function, so I wondered if it's possible after all. I would have thought the inversion required Φ to be 'isolated', and this didn't seem possible here. But I'd be glad to be proven wrong!

    The second one is:

    [itex]N \cdot P(a) = \sum_{i=1}^N {\frac {P(a) \cdot A_i} {P(a) \cdot A_i + [1-P(a)] \cdot B_i} } [/itex]

    where P(a) is as above, N is a positive integer and:

    [itex] A_i = 1 - \Phi \left( \frac {x_i - \mu_a} { \sqrt 2 \cdot \sigma_a} \right) [/itex]

    [itex] B_i = \Phi \left( \frac {x_i - \mu_b} { \sqrt 2 \cdot \sigma_b} \right) [/itex]

    and I need to solve for P(a).

    If I understand correctly, Φ has the property:

    [itex] \Phi (-x) = 1 - \Phi (x) [/itex]

    but I don't see if/how that helps me in this case.

    Any idea?
  2. jcsd
  3. Jun 26, 2016 #2
    It doesn't seem that a numerical analytic solution is possible. And if it is possible, it's likely not really important anyway.
    Last edited: Jun 27, 2016
  4. Jun 27, 2016 #3
    Unless I am much mistaken, a numerical solution is possible, and in fact I did solve both equations numerically for a given dataset of xi's with N ≈ 115000.
    I solved the second one (Excel solver, minimisation of (rhs-lhs)2 ), and with the resulting value of P(a) (≈ 0.33%) I solved the first one (x ≈ 65).

    My question was whether they were analytically solvable.

    Not important, I guess you mean as in 'not a type of problem that is generally encountered in this branch of statistics', as opposed to 'who cares about this' :biggrin:
    If so, I suppose I'd better review the chain of reasoning that led me to these equations.

    PS: just spotted an error in my Tex. The square root of 2 shouldn't be there, it's already in the definition of Φ. Correct version:

    [itex] P(a) \cdot \left[ 1 - \Phi \left( \frac {x - \mu_a} {\sigma_a} \right) \right] = (1 - P(a) ) \cdot \Phi \left( \frac {x-\mu_b} { \sigma_b} \right) [/itex]

    [itex] A_i = 1 - \Phi \left( \frac {x_i - \mu_a} { \sigma_a} \right) [/itex]

    [itex] B_i = \Phi \left( \frac {x_i - \mu_b} { \sigma_b} \right) [/itex]
  5. Jun 27, 2016 #4
    Of course a numerical solution is possible. I made a typo in my post. It doesn't seem an analytic solution is possible. And it's not important anyway, since all we care is a numerical solution.
  6. Jun 27, 2016 #5
    OK, I understand, thanks.

    I would have cared for an analytical solution, because while it's true that I do want the numbers in the end, without a closed form I'll have to write an iteration to solve the problem for each new dataset I get.
    In the past someone else in these forums (@mfb) taught me that I could do an iteration (e.g. Newton) in Excel by just writing out the steps line by line, which worked very well. I'll see if I can do it in this case; much more complicated.
  7. Jun 27, 2016 #6
    Use the following R code:

    Code (Text):

    PA <- 0.5
    mu_a <- 2
    mu_b <- 3
    sigma_a <- 1
    sigma_b <- 2

    f <- function(x) {
      PA*(1 - pnorm((x-mu_a)/sigma_a)) - (1-PA)*pnorm((x-mu_b)/sigma_b)

  8. Jun 28, 2016 #7
    Great, thanks!
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