1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Most probable value given observation

  1. Oct 9, 2014 #1
    Suppose I have observed ##Z = 3##, where ##Z = X + Y##, where ##X \sim N(0,9), Y \sim N(0,4)##. How would I find the most probable value of ##X## that would have given me ##Z = 3##?

    My attempt at a solution: I was given that ##X## and ##Y## are independent, so that means ##Z \sim N(0+0, 9+4) = N(0,13)##. To find the most probable value of ##X##, we would have to find the highest possible probability of ##Z##. But ##Z## is not discrete, so every probability at each point is 0, which means the highest probability of ##Z## is 0. Not sure what I would do after this, so I took a different direction.

    Now I'm trying to find ##E(X|Z=3)## because I would think the mean is the best measure for the "most probable value of X". We have,

    ##E(X|Z=3) = E(X|X+Y=3)##. At this point, are there properties for conditional expectation involving independent random variables to answer this? I tried to find some but wasn't able to. Any help is appreciated.
     
  2. jcsd
  3. Oct 9, 2014 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Yes, but you can still look at the probability density function.

    The expectation value does not have to be the most probable value.
     
  4. Oct 9, 2014 #3

    RUber

    User Avatar
    Homework Helper

    My first instinct is to say that the most likely value of y is 0, so x, having larger standard deviation would be more likely to be 3. However, ##p(x=3 \cap y=0) =.033##. Slightly higher values of y will likely optimize this function.
    in general p(xy)=p(x)*p(y). The probability of a discrete value is not zero.
    upload_2014-10-9_21-23-5.png
     
  5. Oct 9, 2014 #4

    FactChecker

    User Avatar
    Science Advisor
    Gold Member

    Good catch!
    I think it's better to say that the probability density function is not zero. The probability of any single exact number is zero.
     
  6. Oct 9, 2014 #5

    WWGD

    User Avatar
    Science Advisor
    Gold Member

  7. Oct 10, 2014 #6

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Look at
    [tex] f(x|z) \equiv \lim_{\Delta x \to 0, \Delta z \to 0} P(x < X < x + \Delta x\,| z < Z < z + \Delta z)\\
    = \lim_{\Delta x \to 0, \Delta z \to 0} \frac{P(x < X < x + \Delta x \: \cap \: z < Z < z + \Delta z)}{P(z < Z < z + \Delta z)} [/tex]
    For ##z = 3## this will give you
    [tex] c \, f_X(x) f_Y(3-x) [/tex]
    where ##c## is a normalization constant. For what value of ##x## would that be maximized?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Most probable value given observation
Loading...