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Probability Density Function problem

  1. Feb 5, 2017 #1
    1. The problem statement, all variables and given/known data
    Q6. A function, ##f\left(x\right)=\frac{ax+1}{\left(ax-1\right)^3-\frac{a}{\left(x-1\right)^2-1}}##, can be defined as a PDf over the domain ##(0, 2)##.

    Express answers to 4 decimal places unless specified otherwise;
    (a) Find the value of ##a## given that ##f(x)## is a PDf
    (b) Calculate the mean, variance and standard deviation correct to 3 significant figures.
    (c) Determine, correct to one decimal place, the percentage probability of discrete random variable ##x## being within two standard deviations either side of the mean.
    (d) State the value of the median, ##m##
    (e) Find the maximal value of ##f(x)## and determine the percentile of discrete random variable ##x## at this point.
    (f) The derivative of function ##f(x)## can also be defined as a PDf over the domain ##(0, c]##. Find the value of ##c## correct to 3 decimal places, letting ##g(x)=f'(x)##.


    2. Relevant equations
    Knowledge of derivatives, integrals and PDfs.

    3. The attempt at a solution
    The first thing I did was simplify ##f(x)## as I believed it looked rather messy and hard to work with. Sadly I didn't get too far with this - the best I could do was to express it as ##f(x)=\frac{x\left(ax+1\right)\left(x-2\right)}{a\left(3x^3-6x^2-1\right)+\left(x-2\right)\left(a^3x^4-3a^2x^3-x\right)}##

    I then promptly became stuck at the first question. I understand that the question wants me to integrate ##f(x)## using ##0## as the lower limit and ##2## as the upper limit, set whatever my answer is to equal ##1##, and then solve for ##a##. However, I'm not sure how to integrate the function - I don't even think there is a way to calculate the definite integral of this function. Normally this wouldn't be a problem because I'd just use my calculator to integrate, but it's having trouble doing it with two unknowns present. I could substitute in random values of ##a## and then integrate until I get close to the integral equaling ##1##, but that seems like it would be an excessive amount of work and would take a lot of time. Is there another method I could use here to get past this first question? Thank you all for your time and help.
     
  2. jcsd
  3. Feb 5, 2017 #2

    andrewkirk

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    The function is a rational function - meaning a ratio of polynomial functions. Any rational function can be integrated using partial fractions to decompose it into more manageable pieces. For instance see here.

    There may be a neater, easier way to do it, as the partial fractions approach will be somewhat long. But a more elegant approach is not leaping out at me right now.
     
  4. Feb 5, 2017 #3
    Thanks for the advice!

    I'll read up on on partial fraction integration via the link you provided and then give the question another shot.
     
  5. Feb 5, 2017 #4

    Ray Vickson

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    I don't think the partial fraction approach will work in practice; the denominator is a 5th degree polynomial containing ##a## is some of its coefficients, so the roots of the denominator will be functions of ##a##---possibly being unobtainable in closed-form (because it is known to be impossible to give a nice formula for roots of a polynomial of degree >= 5). However, just computing the integral numerically for various numerical values of ##a## works well, especially if you use a computer package such as Maple or Mathematica.
     
  6. Feb 5, 2017 #5
    Hi Saracen Rue:
    I agree with Ray that numerical integration with various trial values for a is the best approach. Here is an online integration site I have used hat might be helpful.
    You may first want to find two values of a such that one gives an integral greater than 1, and the other less than 1. Then keep dividing the reduced range in half. However that may take perhaps 10-20 tries to achieve 4 digits of precision, so some patience will be needed.

    After you have found a value for a that is OK, you can integrate x f(x) and x2 f(x) to calculate mean, variance, and standard deviation.

    Regards,
    Buzz
     
    Last edited: Feb 5, 2017
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