# Probability Density Function problem

1. Apr 23, 2017

### Saracen Rue

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?

2. Apr 23, 2017

### andrewkirk

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.

3. Apr 23, 2017

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.

4. Apr 24, 2017