# Hypothesis testing for log-normal distribution

Assume that we have a random sample from the log-normal distribution and that we want to test $H_0: \tau=0$, where $\tau=\mu+\sigma^2/2$. I want to construct the Wald test for this purpose, and I know that the MLE of $\mu$ and $\sigma^2$ is $\frac{\sum\ln x} {n}$ and $\frac{\sum(\ln x-\hat{\mu})^2} {n}$, respectively. So $$\hat{\tau}=\frac{\sum\ln x} {n}+\frac{\sum(\ln x-\hat{\mu})^2} {2n}$$. The Wald test statistic is defined as $$W=\frac{(\hat{\tau}-\tau_0)^2} {\operatorname{Var}(\hat{\tau})}$$. My question is regarding $\operatorname{Var}(\hat{\tau})$ which I'm having trouble to derive.

• 1. Where did the "$2$" come from in $\hat \tau$? 2. Note that the sample mean and variance of a normal are independent. Can you do the variance of the two parts separately? – Glen_b Jun 1 '17 at 0:59
• Oh the "2" should be in the denominator together with the $n$. Since the second term of $\tau$ is $\sigma^2/2$. Corrected now. – infstat Jun 1 '17 at 4:16

Just in case the image is not visible, the gist is this. Let $Y_i$ denote the log-transformed observations. Let $S^2$ denote the sample variance of the $Y_i$. Then, "an" estimate of the variance of the estimate of $\tau = \mu + \sigma^2/2$ is
$$\widehat{Var}\left(\widehat{\tau}\right) = \dfrac{S^2}{n} + \dfrac{S^4}{2(n-1)}$$
If you are looking for the variance (not an estimate) of $\hat \tau$, you can derive it directly by using the facts that $\bar Y$ and $S^2$ are independent (because if the normality of $Y_i$), $\bar Y$ is normal, and a scalar multiple of $S^2$ is chi-square.
• This looks promising. Is the estimate of the variance of $\hat{\tau}$ the variance of the mle of $\tau$? Because that is important since the Wald test statistic relies on the asymptotic normality of the MLE. – infstat Jun 12 '17 at 21:01
• @infstat: Do you mean: "is the estimate of the variance of $\hat \tau$ the mle of the variance of $\hat \tau$?" If so, on a quick blush it looks like, yes. The variance of $\hat \tau$ (based on the ending paragraph of my answer and math.stackexchange.com/a/72978) is $\frac{\sigma^2}{n} + \frac{1}{4} \frac{2 \sigma^4}{(n-1)}$. So the MLE would be plugging in $S^2$ in place of $\sigma^2$ because of the invariance property of MLE. – Just_to_Answer Jun 16 '17 at 4:14
• A technical correction: The variance of $\hat \tau$ is $\frac{\sigma^2}{n} + \frac{1}{4} \frac{2 \sigma^4}{(n-1)}$. So the MLE would be plugging in $\hat{\sigma^2}$ (the divided by $n$ kind, as opposed to divided by $n-1$ kind unbiased estimate $S^2$) in place of $\sigma^2$ because $\hat{\sigma^2}$ is the MLE of $\sigma^2$. But I don't think divided by $n$ vs. $n-1$ matters for asymptotics (due to Slutsky's). – Just_to_Answer Jun 16 '17 at 4:22