I have a sample:
$4.8, 7.6,1.2, 3.5 ,2.9, 0.8, 0.5, 2.3$ which is to be taking as log normally distributed with params $\mu$ and $\sigma^2$
I am trying to estimate $\mu$ and $\sigma$.
If $X \sim LogN(\mu,\sigma^2)$ then $\log(X) \sim N(\mu,\sigma^2)$. So I try to take the log of each of the sample and the do $\sum(\log(X))/n$ which gives about $0.75$ and I took that as the estimate for $\mu$
However the answers have used the fact that $E(X) = \exp(\mu + 0.5\times\sigma^2)$ and similarly for the $Var(X)$ and solved to get $\mu$ and $\sigma$. I don't understand why these two approaches give different answers
The method of moments works like this: if I have $p$ parameters ($p = 2$ in your case), I must set $$\mathbb{E}[X^k] = \dfrac{1}{n}\sum_{i=1}^{n}x_i^k$$ for $k = 1, 2, \dots, p$.
We have a random variable $X$ which is lognormally distributed with parameters $\mu, \sigma$. Thus, from the method of moments, what you would have to do is set $$\mathbb{E}[X] = \dfrac{1}{n}\sum_{i=1}^{n}x_i = \exp(\hat\mu + \hat\sigma^2/2) = \exp(\hat\mu)\exp(\hat\sigma^2/2)\tag{a}$$ $$\mathbb{E}[X^2] = \dfrac{1}{n}\sum_{i=1}^{n}x_i^2 = \exp(2\hat\mu+2\hat\sigma^2)=\exp(2\hat\mu)\exp(2\hat\sigma^2) \tag{b}$$ Through some algebra, it can be shown that $$\exp(2\hat\mu)\exp(\hat\sigma^2)=\left(\dfrac{1}{n}\sum_{i=1}^{n}x_i\right)^2 $$ (squaring both sides of $(a)$) so thus, taking $(b)/(a)^2$, we obtain $$\exp(\hat\sigma^2)=\dfrac{1}{n}\left(\dfrac{1}{n}\sum_{i=1}^{n}x_i\right)^{-2}\sum_{i=1}^{n}x_i^2 = n\left(\sum_{i=1}^{n}x_i\right)^{-2}\sum_{i=1}^{n}x_i^2$$ or $$\hat\sigma = \sqrt{\ln\left[n\left(\sum_{i=1}^{n}x_i\right)^{-2}\sum_{i=1}^{n}x_i^2 \right]}\text{.}\tag{c}$$ Throwing this into (a), we obtain $$\begin{align} \dfrac{1}{n}\sum_{i=1}^{n}x_i &= \exp\left\{\ln\left[n\left(\sum_{i=1}^{n}x_i\right)^{-2}\sum_{i=1}^{n}x_i^2 \right]/2 \right\}\exp(\hat\mu) \\ &= \exp\ln\left[\sqrt{ n\left(\sum_{i=1}^{n}x_i\right)^{-2}\sum_{i=1}^{n}x_i^2 }\right] \exp(\hat\mu) \\ &= \sqrt{ n\left(\sum_{i=1}^{n}x_i\right)^{-2}\sum_{i=1}^{n}x_i^2 }\exp(\hat\mu) \end{align}$$ or $$\exp(\hat\mu) = \dfrac{1}{n^{3/2}}\left(\sum_{i=1}^{n}x_i^2\right)^{-1/2}\left(\sum_{i=1}^{n}x_i\right)^2$$ or $$\hat\mu = \ln\left[\dfrac{1}{n^{3/2}}\left(\sum_{i=1}^{n}x_i^2\right)^{-1/2}\left(\sum_{i=1}^{n}x_i\right)^2\right]\text{.}\tag{d}$$ Now let $Y$ be the underlying normal random variable of $X$, with $Y = \log(X)$. What you proposed was $$\hat\mu = \dfrac{1}{n}\sum_{i=1}^{n}\ln x_i\tag{e}$$ We can see that (d) and (e) do not agree.