Problem¶
Given \(f(x) = \frac{x}{2}\) for \(0 \leq x \leq 2\). Find the pdf of \(x_1+x_2\) for \(x_1,x_2\) which are i.i.d.
Wrong solution¶
Thus, blindly,
\(f_S(s) = \int_{0}^s \frac{x(s-x)}{4}dx = \frac{s^3}{24}\) and \(0 \leq s \leq 4\)
But is \(f_S(s)\) a PDF?
No.
\(\int_0^4 f_S(s)ds = \frac{4^4}{96} \neq 1\)
Respecting the bounds.¶
\(f_S(s) = \int_{0}^s \frac{x}{2} \frac{s-x}{2}dx\) for :
and
Which gives us, these bounds:
\(s-2 \leq x \leq s\) and \(0 \leq x \leq 2\)
In order to respect the above bounds,
\(f*g(s) = \int_{-\infty}^{\infty}f(x)f(s-x)dx\)
But, \(f(x) \neq 0\) for \(0 \leq x \leq 2\)
and \(f(s-x) \neq 0\) for \(s-2 \leq x \leq s\)
thus, \(f(x)f(s-x) \neq 0\) for \(max(0,s-2) \leq x \leq min(s,2)\)
Range of s: \([0,4]\)
\(f*g(s) = \int_{max(0,s-2)}^{min(s,2)}f(x)f(s-x)dx\)
For \(0 \leq s \leq 2\) : \(min(s,2)=s\) and \(max(0,s-2)=0\)
For \(2 \leq s \leq 4\) : \(min(s,2)=2\) and \(max(0,s-2)=s-2\)
Now,