Skip to main content
added 37 characters in body
Source Link
whuber
  • 333.5k
  • 63
  • 792
  • 1.3k

You can use the inverse erf function, which is available in MatLab and Mathematica, for instance.

For the normal CDF, starting from

$$y=\Phi\left(x\right)=\frac{1}{2}\left[1+erf\left(\frac{x}{\sqrt{2}}\right)\right]$$$$y=\Phi\left(x\right)=\frac{1}{2}\left[1+\text{erf}\left(\frac{x}{\sqrt{2}}\right)\right]$$

We get

$$x=\sqrt{2}erf^{-1}\left(2y-1\right)$$$$x=\sqrt{2}\ \text{erf}^{-1}\left(2y-1\right)$$

For the log-normal CDF, starting from

$$y=F_{x}(x;\mu,\sigma)=\frac{1}{2}erfc\left(\frac{-lnx-\mu}{\sigma\sqrt{2}}\right)$$$$y=F_{x}(x;\mu,\sigma)=\frac{1}{2}\text{erfc}\left(\frac{-\log x-\mu}{\sigma\sqrt{2}}\right)$$

We get

$$-\ln\left(x\right)=\mu+\sigma\sqrt{2}erfc^{-1}\left(2y\right)$$$$-\log \left(x\right)=\mu+\sigma\sqrt{2}\ \text{erfc}^{-1}\left(2y\right)$$

You can use the inverse erf function, which is available in MatLab and Mathematica, for instance.

For the normal CDF, starting from

$$y=\Phi\left(x\right)=\frac{1}{2}\left[1+erf\left(\frac{x}{\sqrt{2}}\right)\right]$$

We get

$$x=\sqrt{2}erf^{-1}\left(2y-1\right)$$

For the log-normal CDF, starting from

$$y=F_{x}(x;\mu,\sigma)=\frac{1}{2}erfc\left(\frac{-lnx-\mu}{\sigma\sqrt{2}}\right)$$

We get

$$-\ln\left(x\right)=\mu+\sigma\sqrt{2}erfc^{-1}\left(2y\right)$$

You can use the inverse erf function, which is available in MatLab and Mathematica, for instance.

For the normal CDF, starting from

$$y=\Phi\left(x\right)=\frac{1}{2}\left[1+\text{erf}\left(\frac{x}{\sqrt{2}}\right)\right]$$

We get

$$x=\sqrt{2}\ \text{erf}^{-1}\left(2y-1\right)$$

For the log-normal CDF, starting from

$$y=F_{x}(x;\mu,\sigma)=\frac{1}{2}\text{erfc}\left(\frac{-\log x-\mu}{\sigma\sqrt{2}}\right)$$

We get

$$-\log \left(x\right)=\mu+\sigma\sqrt{2}\ \text{erfc}^{-1}\left(2y\right)$$

Post Merged (destination) from stats.stackexchange.com/questions/22836/…
I detailed my argument in order to conform to a suggestion from one of my commenters.
Source Link

Maybe you couldYou can use the inverse erf function?, which is available in MatLab and Mathematica, for instance.

For the normal CDF, starting from

$$y=\Phi\left(x\right)=\frac{1}{2}\left[1+erf\left(\frac{x}{\sqrt{2}}\right)\right]$$

We get

$$x=\sqrt{2}erf^{-1}\left(2y-1\right)$$

For the log-normal CDF, starting from

$$y=F_{x}(x;\mu,\sigma)=\frac{1}{2}erfc\left(\frac{-lnx-\mu}{\sigma\sqrt{2}}\right)$$

We get

$$-\ln\left(x\right)=\mu+\sigma\sqrt{2}erfc^{-1}\left(2y\right)$$

Maybe you could use the inverse erf function?

You can use the inverse erf function, which is available in MatLab and Mathematica, for instance.

For the normal CDF, starting from

$$y=\Phi\left(x\right)=\frac{1}{2}\left[1+erf\left(\frac{x}{\sqrt{2}}\right)\right]$$

We get

$$x=\sqrt{2}erf^{-1}\left(2y-1\right)$$

For the log-normal CDF, starting from

$$y=F_{x}(x;\mu,\sigma)=\frac{1}{2}erfc\left(\frac{-lnx-\mu}{\sigma\sqrt{2}}\right)$$

We get

$$-\ln\left(x\right)=\mu+\sigma\sqrt{2}erfc^{-1}\left(2y\right)$$

Source Link

Maybe you could use the inverse erf function?