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Homework Help: Mixed random variable distribution question

  1. Oct 25, 2017 #1
    1. The problem statement, all variables and given/known data
    See attached image (See below)

    2. Relevant equations
    Differential equations.
    And a combination of discrete & continuous distributions

    3. The attempt at a solution
    The Continous Distribution Function (CDF) is given in the question. So I differentiated it with respect to x piecewise (See below). That is what I got

    I can also see that there is a uniform distribution between 0 and 1 but also there are discrete points that have assigned probability. For example; those points could be 0,1 and 2? maybe?
    I want to be able to find the value of c while remembering the sum of the probabilities must add up and equal to 1.

    How do I go about solving for c? this is where I am stuck. I cannot seem to figure out where to go from here. Thanks!

    Question:
    2b.png Differentiated with respect to xpiecewise.png (When I differentiated it)
     
  2. jcsd
  3. Oct 25, 2017 #2

    andrewkirk

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    The calculation of the density function ##f## is not quite correct. The density is infinite at ##x=0,2## and also possibly at 1 (depending on the value of ##c##), because the CDF jumps up discontinuously at those points. We say there is a 'discrete probability mass' at those infinite-density points, meaning there is a non-zero probability of X having exactly that value.

    You need to use the additional fact that ##E[X]=1##. Using the formulas you got for ##f##, corrected to also include the points of discrete probability mass at 0 and 2, write an expression for ##E[X]## in terms of ##c##. Equating that expression to 1 will give you an equation you can solve to find ##c##. The expression will be a sum of integrals and isolated discrete terms.

    There are two cases to consider here - (1) where there is a discrete probability mass at ##x=1## and (2) where there is not. I suggest trying case (1) first. If that does not give an acceptable solution, try case (2).
     
  4. Oct 25, 2017 #3
    Thank you for that. I figured it might have had something to do with the expectation formula. E[X] = 1 in terms of c. There are definitions for the expectation of continuous random variables and for discrete ones as well. For a mixed distribution you would need to mix both definitions together right?
    I'm having a lot of trouble trying to figure out how to mix these two expectation formulas
     
  5. Oct 25, 2017 #4

    andrewkirk

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    Just integrate ##xf(x)## over the domain excluding the discrete mass points, and add the sum across each of those points of (probability of landing on the point) x (value of x at that point).
     
  6. Oct 25, 2017 #5
    This is what I did but I don't think I'm going in a correct direction. C = constant whereas c = the value i'm looking for. 1.png
     
  7. Oct 25, 2017 #6

    Ray Vickson

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    For a non-negative random variable ##X##, whether continuous, discrete or mixed, if ##F(x) = P(X \leq x)## is the CDF and ##G(x) = P(X > x) = 1 - F(x)## is the complementary CDF, then we have that
    $$ EX = \int_0^{\infty} G(x) \, dx $$
    You can do the integral in terms of ##c## and so get an equation for ##c##.
     
  8. Oct 25, 2017 #7
    I got a c value of 1/3 by doing this (1 = c/2 + ½ -c + 2/3 )
    Then for part (i): Probability = 1/6th or 16.67%
    (ii) Probability = 2/3rds or 66.67%
    (iii) Probability = 1/3rd or 33.3%
    (iv) Probability = 1/2 or 50%

    I think I've got it now providing those are the right answers!
     
  9. Oct 25, 2017 #8

    Ray Vickson

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    The first two are; I have not checked the other two.

    However, in English, "1/6" is already pronounced as "one-sixth", so when you write 1/6th you are really writing "one-sixth-th". Thus, the The first two answers are just 1/6 and 2/3 (not "two-thirds-irds").
     
    Last edited: Oct 25, 2017
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