# How to differentiate the distribution function of lognormal distribution with respect to its parameters?

How to differentiate the distribution function of lognormal distribution with respect to its parameters? What solution will we get? I know if differentiate wrt variable, we will get density function.!

• Please present your notation. For the standard notation you will find the equations at mathworld.wolfram.com. Commented Aug 12, 2017 at 10:42
• @Semoi yes I have added now..Sorry i dont know how to write equations in this page.. so I have uploaded image of equations. Commented Aug 12, 2017 at 12:28
• @Huchesh what is the context for your question?
– user137329
Commented Aug 12, 2017 at 12:52
• @phdmba7of12 actually i am using truncated lognormal distribution (TLD). In truncated distribution (density of the distribution divided by the distribution function of the distribution at the specific point) , denominator is distribution function which is in the form of cumulative distribution function in case of lognormal distribution. And i want to estimate the parameters of the LTD but due to the not explicit form of likelihood equations (LE's), i have to use numerical methods like newton raphson method. In NR method, i have to take the derivatives of LE's. conti.. Commented Aug 15, 2017 at 3:54
• @phdmba7of12 So i am unable to differentiate that distribution function with respect to parameters. I hope, u got my point. Commented Aug 15, 2017 at 3:55

Well, as you stated the cumulative distribution function is $$F(x) = \int_{-\infty}^x f(z) dz = \Phi\big[\frac{\ln(x)-\mu}{\sqrt{2}\sigma}\big]$$
Now I would differentiate the error fct using the chain rule $$\frac{\partial}{\partial y}\Phi[z] = \frac{\partial \Phi[z]}{\partial z} \cdot \frac{\partial z}{\partial y} = f(z) \cdot \frac{\partial z}{\partial y}$$ So all you need to do is to calculate the partial derivatives \begin{align} \frac{\partial }{\partial \mu} \big[\frac{\ln(x)-\mu}{\sqrt{2}\sigma}\big] &= ... \\ \frac{\partial }{\partial \sigma} \big[\frac{\ln(x)-\mu}{\sqrt{2}\sigma}\big] &= ... \end{align} which is straight forward.
• Well, $\frac{\partial}{\partial \mu} \mu = 1$ and $\frac{\partial}{\partial \sigma} \sigma^{-1} = -\sigma^{-2}$. Commented Aug 15, 2017 at 17:31