# Match Right Skewed Distribution to Normal

I am running a simulation. One of my parameters is sampled from a normal distribution. I would like to perform a sensitivity analysis using a right skewed distribution.

This is what I had hoped to do: specify a log normal distribution, with a leftward translation so that it has the same cumulative density below zero as the reference normal, and have the same mean and variance (on the log scale) as the reference normal.

I have played around with these parameters. I am stuck. 1) this is even possible? 2) if it is possible, I may need to solve for the parameters numerically... But I am having trouble specifying an equation(s). I am having trouble with the algebra

If it is impossible to have the exact same expectation, variance, and cumulative density below zero, I'd be happy with letting the variance be unconsteained.

Alternatively, I'd be happy using any right skewed distribution, I only started with the log normal because I thought it would be simple.

I can post the equations I have worked on, but I suspect they are not useful. I've been using the definitions posted at this link (http://www.mathworks.com/help/stats/lognpdf.html)

Let $X\sim N(\mu_X,\sigma_X^2)$ and $F_X(0) = \Phi(-\mu_X/\sigma_X)$.

Let $Y\sim\text{logN}(\mu, \sigma^2)$.

Let $Y_\gamma\sim\text{shifted-logN}(\mu, \sigma^2,\gamma)$.

In what I do here $\gamma$ is the left-shift, so positive $\gamma$ means its origin is shifted below 0.

$\text{mean}= \exp(\mu+\frac{_1}{^2} \sigma^2)-\gamma$

$\text{variance}= \exp(\mu+\frac{_1}{^2} \sigma^2)^2[\exp(\sigma^2)-1]$

$P(Y-k<0)=P(Y<k) = P(\log(Y)<\log(k))= \Phi[(\log(k)-\mu)/\sigma]$

$F_\gamma(0) = \Phi[\frac{\log(\gamma)-\mu}{\sigma}]$.

That is:

$\Phi((\log(\gamma)-\mu)/\sigma)=\Phi(-\mu_X/\sigma_X)$.

So:

$(\log(\gamma)-\mu)/\sigma=-\mu_X/\sigma_X\hspace{4.5cm}(1)$

$\exp(\mu+\frac{_1}{^2} \sigma^2)-\gamma = \mu_X\hspace{5cm}(2)$

$\exp(2\mu +\sigma^2)[\exp(\sigma^2)-1]=\sigma^2_X\hspace{3.35cm}(3)$

Some manipulation leads to a complicated expression relating $\sigma$, $\mu_X$ and $\sigma_X$:

$$\exp(\frac{_1}{^2}\sigma^2)\sqrt{\exp(\sigma^2)-1}=C_X\left[\exp(\frac{_1}{^2}\sigma^2)-\exp(-\sigma/C_X)\right]$$

where $C_x = \sigma_X/\mu_X$, the coefficient of variation. It can be simplified a bit further.

- I don't think this is solvable in closed form algebraically, but it could be solved numerically (by root-finding); a good starting point or points might be possible to obtain via series expansion.

[If you can't get this out, I'll come back with a derivation later.]

Then simple expressions can be used to obtain $\gamma$ and $\mu$:

$$\exp(\mu)=\frac{\sigma_X}{\exp(\frac{_1}{^2}\sigma^2)\sqrt{\exp(\sigma^2)-1}}$$

$$\gamma = \exp(\mu+\frac{_1}{^2}\sigma^2)-\mu_X$$

So I don't think this is possible to solve this problem as it reduces to a sum of exponentials. So what I did was generate a number of lognormal distributions, and solve for a normal matching the above criteria