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Probability Density Function problem

  1. Apr 23, 2017 #1
    1. The problem statement, all variables and given/known data
    Presume the relation ##\frac{x}{x+y^2}-y=x## is defined over the domain ##[0,1]##.

    (a) Rearrange this relation for ##y## and define it as a function, ##f(x)##.
    (b) Function ##f(x)## is dilated by a factor of ##a## from the y-axis, transforming it into a probability density function, ##p(x)##. Find the value of ##a## correct to 4 decimal places.
    (c) Determine the following correct to 3 decimal places:

    I) The mean of ##p(x)##
    II) The standard deviation of ##p(x)##
    III) The median, ##m##, of ##p(x)##
    (d) Calculate the probability of discrete random variable ##x## being within ##a## standard deviations either side of the mean.

    2. Relevant equations
    Knowledge of integration, probability density functions, and the rearranging and solving of equations.

    3. The attempt at a solution
    Starting with part ##(a)##, I attempted to rearrange ##\frac{x}{x+y^2}-y=x## for ##y##. I managed to express the equation in the form ##y^3+xy^2+xy+x^2-x=0## however this is where I become stuck. I'm unsure of how to factorise this equation for ##y## and my calculator simply returns an error message when I try and use it. Is there another way to do this that I'm missing or don't know about?

    Thank you for taking your time to read this :)
     
  2. jcsd
  3. Apr 23, 2017 #2

    andrewkirk

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    Your calculations so far look correct. I don't think that formula can be factorised. A formula for the solution of the cubic in y could be written (see cubic solution formula here) but it would be very messy and I doubt that's what was intended.
    Perhaps the question contains an error like a wrong sign, and the question it was supposed to be is factorisable where the one they actually wrote is not.

    Where did you get the question? It has a number of other errors, such as
    • the statement that the relation is over the domain [0,1], which cannot be correct as there are two variables and [0,1] is only one-dimensional. Perhaps they meant to say [0,1] x [0,1].
    • the references to the mean, standard deviation and median of p(x) are meaningless, since those statistics are properties of random variables and p(x) is not a random variable. Perhaps they meant to say the random variable whose pdf is p.
     
  4. Apr 23, 2017 #3

    Charles Link

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    There may be a simple way to factor it, but with a little effort in trying to factor it, I came up empty. Suggestion would be to do the formal solution of the cubic equation on it in order to determine the factors.
     
  5. Apr 24, 2017 #4

    Charles Link

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    A follow-on: I worked through most of the formal solution of the cubic equation for this problem. Unless I made algebraic errors, it doesn't appear to simplify a great deal and you get complicated polynomials of powers of x from the 5th power to the second power (the 6th power term cancelled) inside of a square root sign. That part is the solution to a quadratic equation that you add an expression consisting of a 3rd power polynomial in x and then take a cube root of it. Finally, you would then do a similar computation to get the "s" term, (solving for s and t), y'=s-t, and then y=y'-x/3. This one does not appear to be simple.
     
  6. May 3, 2017 #5

    haruspex

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    Maybe you are not expected to get it into the form y=.... Try proceeding to the next part, which involves computing ∫y.dx. Can you see a way to do that?
     
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