MHB Cumulative Distribution function in terms of Error function

[TheFallen018]

Hey guys,

I've got this problem I've been trying to solve, but it makes little sense to me. I've tried a few things, but I feel like with each method, I've made no progress, and I haven't been able to make the problem make any more sense to me by trying those things.

Here's the question:

The error function is \begin{align*}\frac{1}{\sqrt{\pi}}\int_{-x}^{x}e^{-t^2} dt\end{align*}
The cumulative distribution function for $x\geq0$ is \begin{align*}\frac{1}{2}+\frac{1}{\sqrt{2\pi}}\int_{0}^{x}e^-\frac{t^2}{2} dt\end{align*}

By making a suitable substitution, find a formula for the cumulative distribution function in terms of the error function when $x\geq0$

The previous question had to do with taking the derivative with respect to x of the error function, so I was thinking I could try integrating that again and use parts of that, but that obviously didn't work. I've lost track of where my thoughts are, so I was hoping someone would be able to point me in the right direction.

Thank you :)

 

[GJA]

Hi TheFallen018,

Here are a few hints:

1. Notice that the error function is an even function. Use this fact to find an equivalent integral form for the error function.
2. Make a substitution in your cumulative distribution function so that $e^{-\frac{t^{2}}{2}}$ becomes $e^{-t^{2}}$ to match the form of the exponential in the error function.
3. Do a little algebra to combine your answers from Steps 1 & 2 to get an expression for the distribution in terms of the error function.

Let me know if anything requires further clarification.