# Partial derivative of the likelihood function respect to $\sigma^2$

I am having problem doing the partial derivative of the likelihood function which is

$$L(\mu,\sigma^2)=\frac{1}{\sigma\sqrt{2\pi}^n}\times \exp{(-\frac{1}{2\sigma^2}\sum(x_i-\mu)^2)}$$

If the first part has solved that the $$\hat{\mu}$$ is $$\bar{x}$$ and plug this to the $$L(x,\mu,\sigma)$$ I wonder may I ask how to make the partial derivative to the $$\sigma^2$$ and the answer is $$-n/2\sigma^2+1/2\sigma^4\times\sum(x_i-\mu)^2$$

Thank you! I guess I did something wrong so I tried two times but did not get the above answer.

Appreciated!

We generally work with the log likelihood. Like the log function is increasing, the maximum of the likelihood is also the maximum of the log likelihood.

$$\log{L(\mu,\sigma^2)}=-\frac{n}{2}\log{\sigma^2}-\frac{1}{2\sigma^2}\sum\limits_{i=1}^n(x_i-\mu)^2+C$$

where $$C$$ is a constant term (that does not depend on $$\sigma$$ or $$\mu$$). This constant is generally dropped because it does not play any role in our maximization.

Also note that in order to express everything in $$\sigma^2$$ (and not in $$\sigma$$) I have used this "trick": $$\log{\sigma}=\log{(\sigma^2)^{1/2}}=\frac{1}{2}\log{\sigma^2}$$ (valid because we assume $$\sigma>0$$)

Now you can compute your partial derivatives:

$$\frac{\partial}{\partial \mu}(\log{L(\mu,\sigma^2)})=\frac{1}{\sigma^2}\sum _{i=1}^n \left(x_i-\mu \right)$$

$$\frac{\partial}{\partial \sigma^2}(\log{L(\mu,\sigma^2)})=\frac{\sum _{i=1}^n \left(x_i-\mu \right)^2}{2 \sigma ^4}-\frac{n}{2 \sigma ^2}$$

Next to get $$(\mu,\sigma^2)$$ you must solve $$\nabla\log{L(\mu,\sigma^2)}=\mathbf{0}$$, that is:

$$\sum _{i=1}^n \left(x_i-\mu \right) = 0$$

$$\sum _{i=1}^n \left(x_i-\mu \right)^2=n\sigma^2$$

($$\mu=\bar{x}$$ and $$\sigma^2=\frac{1}{n}\sum _{i=1}^n \left(x_i-\bar{x}\right)^2$$)

update (see comment): clarification concerning $$\frac{\partial}{\partial \sigma^2}(-\frac{1}{2\sigma^2})=\frac{1}{2\sigma^4}$$ which can be a little confusing.

You must realize that "$$\sigma^2$$" must be interpreted purely as a symbol. The previous calculation must be thought as follows:

In order to compute $$\frac{\partial}{\partial \sigma^2}(-\frac{1}{2\sigma^2})$$ what you actually do is to replace the "symbol" $$\sigma^2$$ by $$x$$: $$\frac{\partial}{\partial x}(-\frac{1}{2x}) = \frac{1}{2x^2}$$ then you reintroduce $$\sigma^2$$, the complete story is: $$\frac{\partial}{\partial \sigma^2}(-\frac{1}{2\sigma^2}) \equiv \frac{\partial}{\partial x}(-\frac{1}{2x}) = \frac{1}{2x^2} \equiv \frac{1}{2(\sigma^2)^2}= \frac{1}{2\sigma^4}$$

Another example (assuming that $$\sigma>0$$) would be: $$\frac{\partial}{\partial \sigma^2}\frac{1}{\sigma}=\frac{\partial}{\partial \sigma^2}\frac{1}{\sqrt{\sigma^2}}\equiv\frac{\partial}{\partial x}\frac{1}{\sqrt{x}}=-\frac{1}{2 x^{3/2}}\equiv -\frac{1}{2 (\sigma^2)^{3/2}}=-\frac{1}{2\sigma^3}$$

• Very clear, thank you so much!!
– Chen
Nov 14 '18 at 21:01
• @Chen happy if it helps :) Nov 14 '18 at 21:02
• @Vincent, may I ask how did you come up with the $\sigma^4$? Mine has been $\sigma^3$ in the denominator. Thanks!!
– Chen
Nov 14 '18 at 21:17
• @chen see my update, does it answer your question? Nov 14 '18 at 21:48
• Thank you so much @Vincent, this is really helpful for me!!!
– Chen
Nov 16 '18 at 15:24