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Probability Density Function

  1. Nov 20, 2007 #1
    Q: Given f(x) = cx + (c^2)(x^2), 0<x<1.
    What is c such that the above is a proper probability density function?

    ∫ f(x) dx = 1
    => 2(c^2) + 3c - 6 =0
    => c= (-3 + sqrt57) / 4 or c= (-3 - sqrt57) / 4
    => Answer: c= (-3 + sqrt57) / 4 (the second one rejected)

    Now, what is the reason of rejecting c= (-3 - sqrt57) / 4 ?

    Also, in which step is this extraneous solution produced and why is it produced?

    Thanks for explaining!
  2. jcsd
  3. Nov 20, 2007 #2


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    [itex]\int_0^1 cx+ c^2x^2 dx= c/2+ c^2/3[/itex]. Setting that equal to 1, c/2+ c2/3= 1, and multiplying through by 6, 3c+ 2c2= 6 or 2c2+ 3c- 6= 0. That has, as you say, roots of
    [tex]c= \frac{-3\pm\sqrt{57}}{2}[/tex]

    Because the cumulative probabilty must be increasing, if c were negative we could have a section on which the probability from a to b (b> a) might be negative and that is impossible. As to "where the extraneous solution was produced", it was really produced when the "c2" was put into the problem. That introduced the possiblity of a negative value of c.
    Last edited: Nov 20, 2007
  4. Nov 20, 2007 #3
    I don't see why the second one is rejected. Does it say it's rejected or is it just not in the answer?
  5. Nov 20, 2007 #4
    Ah, OK. Nevermind. Makes sense.
  6. Nov 20, 2007 #5
    Graphically speaking, if you draw f(x) and notice a section of it below the x-axis, then there's chance of getting a negative probability in some interval [a, b] which is a no no. That means that f must satisfy that additional property that f(x) >= 0 for all values of x. Right?
  7. Nov 20, 2007 #6
    OK, but how do you know for sure that for the second "c", there IS a section on which the density function is negative? (without a graphing calculator since I am not allowed to use it...)

    But the original density function has a c^2 too, I did NOT "square both sides", so there shouldn't be any extraneous solution, right?

    Thanks a lot for your help and explanation!
  8. Nov 21, 2007 #7


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    If c= (-3-[itex]\sqrt{57})/4, which is approximately -2.7, then your probability density function is f(x) = cx + (c^2)(x^2) or f(x)= 6.9x2- 2.7x. That's a parabola and has zeros at x= 0 and x= 2.7/6.9= .391! If x= .25, f(.25)= 6.9(.0625)- 2.7(.25)= -.632> 0 which means that f(x)< 0 for all x between 0 and .391. If you were to use that value for c, you would find that the "probability" that x las between, say 0 and .25 was negative!
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