# Difficult Probability Density Function Question

1. Sep 25, 2016

### Saracen Rue

1. The problem statement, all variables and given/known data
A function, $f\left(x\right)=\left|a^{\frac{\sin \left(x\right)}{\ln \left(ax\right)}}-\frac{x}{a}\right|$, intersects with another function, $g\left(x\right)=\left|\frac{sin(ln(\sqrt{x}-\sqrt{a}))}{x^2-a^2}\right|$, at point $Q(b,f(b))$ and point $R(c,f(c))$. A probability density function, $p(x)=g(x)-f(x)$, can be formed over the domain $[b,c]$, where $0<b<c<2$.

Q1. Determine, correct to 16 decimal places;
(a) The value of the real-valued constant, $a$
(b) The coordinates of the points $Q$ and $R$
Q2. Calculate the mean, variance and standard deviation correct to 4 decimal places

Q3. Find correct to 1 decimal place; the percentage probability of the continuous random variable, $X$, being within $a^2$ variances either side of the mean (i.e. $Pr(μ-a^2⋅Var(X)≤X≤μ+a^2⋅Var(X))$)

2. Relevant equations
For PDfs: $∫_{b}^{c}f(x)dx=1$

3. The attempt at a solution
I know that to find points $b$ and $c$ I need to solve $f(x)=g(x)$ for $x$, however when I try to do this on my calculator it gives me an error message. I know that I should be getting $b$ and $c$ in terms of $a$, thus $∫_{b}^{c}p(x)dx=1$ should have $a$ as the only unknown and it should be able to be solved for. After getting $a$, I'd substitute the value into $b$ and $c$ to get those values as well, and then I'd be able to find $f(b)$ and $f(c)$ as well (giving me points $Q$ and $R$).

To find the mean, variance and standard deviation I'd just solve the following:
$E(X)=∫_{b}^{c}x⋅p(x)dx$, $Var(X)=∫_{b}^{c}x^2⋅p(x)dx$ and $σ=\sqrt{Var(X)}$

And for $Q3$ I'd solve this: $Pr(μ-a^2⋅Var(X)≤X≤μ+a^2⋅Var(X))=∫_{μ-a^2⋅Var(X)}^{μ+a^2⋅Var(X)}p(x)dx$

I am fairly confident with my method for the most part (although I'd still appreciate somebody telling me if there's any errors in it), the main issue I'm having is simply not being able to solve the equations on my calculator. I'd be very thankful for anybody who has an alternative way for me to solve this problem :)

2. Sep 25, 2016

### andrewkirk

If we can identify a nice search space then it shouldn't be too hard to numerically find a suitable value of $a$.

To start that process, we can observe that, if we want $f(x)-g(x)$ to be defined and continuous on $b,c$, we must have $1\leq a<b$. Can you see why that must be the case?

Given that, we can differentiate $h\triangleq f-g$ and set it to zero to try to find extrema, of which there will have to be at least one between $b$ and $c$.

Or - quicker and easier if it's allowed, just write a short bit of code in your favourite number-crunching language to search for a value of $a\in[1,2]$ for which $h(x)$ attains a maximum or minimum in the interval $[a,2]$ and then for such an $a$ see if the curve of $h$ crosses the $x$ axis on either side of that extremum.

By the way, the question is worded in a way that seems to imply that there is only one possible value of $a$. In fact there is a range of possible values. So you only need to find a value of $a$ that will work.

3. Sep 26, 2016

### Ray Vickson

I think the question is implying that there will be only one value of $a$ for which the equation $1 = \int_{b(a)}^{c(a)} h(x,a) \, dx$ will work.