Cumulative Distribution function in terms of Error function

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SUMMARY

The discussion focuses on deriving the cumulative distribution function (CDF) in terms of the error function (erf) for the standard normal distribution. The error function is defined as \(\frac{1}{\sqrt{\pi}}\int_{-x}^{x}e^{-t^2} dt\), while the CDF for \(x \geq 0\) is given by \(\frac{1}{2}+\frac{1}{\sqrt{2\pi}}\int_{0}^{x}e^{-\frac{t^2}{2}} dt\). Key steps include recognizing the even nature of the error function and substituting variables to align the exponential terms. The final expression for the CDF can be derived by combining these insights through algebraic manipulation.

PREREQUISITES
  • Understanding of the error function (erf)
  • Familiarity with cumulative distribution functions (CDF)
  • Knowledge of integration techniques in calculus
  • Ability to perform variable substitutions in integrals
NEXT STEPS
  • Study the properties of the error function and its applications in statistics
  • Learn about the derivation of cumulative distribution functions from probability density functions
  • Explore integration techniques, particularly variable substitution methods
  • Investigate the relationship between the normal distribution and the error function
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Mathematicians, statisticians, and students studying probability theory who are looking to deepen their understanding of the relationship between the error function and cumulative distribution functions.

TheFallen018
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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 :)
 
Last edited:
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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.
 
Last edited:

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