I am struggling with a hypothesis test between the mean and variance of a sample of i.i.d Gaussian random variables. This (self-study) question arises in the context of the Delta Method (first or second order) and constructing asymptotic test statistics:

Based on a sample of i.i.d. Gaussian random variables $X_1, . . . , X_n$ with mean $\mu$ and variance $\sigma^2$, propose a test with asymptotic level 5% for the hypotheses

$H_0 : \mu >\sigma \ vs \ H_1 : \mu \leq \sigma$

Potential hints from the lecture notes:

Given a hypothesis of the form: $H_0: g(\theta) = 0$ where $g: \mathbb{R^d} \to \mathbb{R^k}$

We are given:

  • If the MLE estimator is asymptotically normal: $\sqrt{n} (\hat \theta - \theta) \xrightarrow{d} N_d(0,\Sigma(\theta))$
  • The first order delta method gives: $\sqrt{n} (g(\hat \theta) - g(\theta)) \xrightarrow{d} N_k(0,\Gamma(\theta))$ which can be rewritten as $\sqrt{n} \ \ \Gamma(\theta)^{-1/2} (g(\hat \theta) - g(\theta)) \xrightarrow{d} N_k(0,I_k)$
    • $\Gamma (\theta) = \nabla g(\theta)^T \Sigma(\theta)\nabla g(\theta) \in \mathbb{R}^{k \times k}$
    • Assume $\Sigma$ is invertible, $\nabla g$ has rank $k$, $\Gamma$ is invertible and continuous in $\theta$
  • By Slutsky's theorem we can replace $\theta$ with the MLE $\hat \theta$ such that $\sqrt{n} \ \Gamma (\hat \theta)^{-1/2} \ (g(\hat \theta) - g(\theta)) \xrightarrow{d} N_k(0,I_k)$
  • Second order delta method: $n \ g(\hat \theta)^T \ \Gamma \ (\hat \theta)^{-1} g(\theta) \ \xrightarrow{d} \chi^2_k$

First attempt

We can reformulate the null hypothesis as: $H_0 : g(\mu, \sigma): \mu - \sqrt{\sigma^2} < 0$ which maps from $\mathbb{R^2} \to \mathbb{R}$ so $k = 1$

\begin{aligned} & \frac{\partial g}{\partial \mu} = 1 \\ & \frac{\partial g}{\partial \sigma^2} = - \frac{1}{2\sigma} \\ & \nabla g(\mu, \sigma) = [1,- \frac{1}{2\sigma}]^T \end{aligned}

The MLE estimators for the normal distribution are:

  • $\hat \mu = \bar{X}_n$ with asymptotic distribution $\xrightarrow{d} N(\mu, \frac{\sigma^2}{n})$
  • $\hat \sigma^2 = \frac{1}{n} \sum X_i - \bar{X}_n^2$ with asymptotic distribution $\xrightarrow{d} N(\sigma^2, \frac{2\sigma^4}{n})$

The covariance matrix $\Sigma$ for the sample of i.i.d. normal r.v. can be found from the Fisher Matrix or otherwise:

$\Sigma(\mu, \sigma^2) = \begin{bmatrix}\frac{\sigma^2}{n} & 0 \\0 & \frac{2 \sigma^4}{n} \end{bmatrix}$

Computing $\Gamma$ which should be a $1 \times 1$ matrix as $k =1$ \begin{aligned} \Gamma &= \nabla g^T \Sigma \nabla g \\ & = [1,- \frac{1}{2\sigma}] \ \begin{bmatrix}\frac{\sigma^2}{n} & 0 \\0 & \frac{2 \sigma^4}{n} \end{bmatrix} \ [1,- \frac{1}{2\sigma}]^T \\ & = \frac{3 \sigma^2}{n} \\ \Gamma^{-1/2} & = \frac{\sqrt{n}}{\sigma \sqrt{3}} \end{aligned}

Computing the test statistic based on the first order Delta Method (i.e. normal distribution) assuming that under $H_0 : g(\mu, \sigma) = 0$ (How can we deal with the inequality ?)

\begin{aligned} T_n & = \sqrt{n} \ \ \Gamma(\theta)^{-1/2} (g(\hat \theta) - g(\theta)) \\ & = \sqrt{n} \ \frac{\sqrt{n}}{\hat \sigma \sqrt{3}} [ (\hat \mu - \hat \sigma) - 0 ] \\ & = \ \frac{n (\hat \mu - \hat \sigma)}{\hat \sigma \sqrt{3}} \end{aligned}

Computing the second order delta method test statistic (i.e. chi-squared distribution)

\begin{aligned} T_n & = n \ \ g(\hat \theta)\ \ \Gamma(\theta)^{-1}\ g(\hat \theta) \\ & = n (\hat \mu - \hat \sigma) \frac{n}{3 \sigma^2} (\hat \mu - \hat \sigma) \\ & = \frac{n^2 (\hat \mu - \hat \sigma)}{3 \hat \sigma^2} \end{aligned}


  • Is this the right way to deal with such a hypothesis test ?
  • What distinguishes the first order test statistic (normal distribution) from the second order test statistic (Chi-squared distribution)
  • What happens to the inequality sign ?
  • $\begingroup$ I think, what you did is right. The delta method must me second order since there are two unknown parameters, know just test the hypothesis. $\endgroup$ – user13 Apr 28 '18 at 17:50
  • $\begingroup$ However, I am not sure why you calculated the derivative with respect to $σ^2$ rather than σ, because your parametr is σ, isn't it? $\endgroup$ – user13 Apr 28 '18 at 17:59
  • $\begingroup$ In fact I had first done this as you said, by taking the derivative w.r.t $\sigma$ rather than $\sigma^2$ - but this then required to apply the Delta method directly to $\sigma$ in order to find its distribution and variance. The result was the same, but the derivation seemed less elegant than this one which uses matrix multiplication. $\endgroup$ – Xavier Bourret Sicotte Apr 29 '18 at 9:59
  • $\begingroup$ I found a calculation mistake. Asymptotic variance should be 3(sigma^2)/2. $\endgroup$ – ROHIT GUPTA Apr 16 at 13:45
  • $\begingroup$ Really? Please post you calculations as an answer! $\endgroup$ – Xavier Bourret Sicotte Apr 16 at 15:03

Few notes:

1)In last eqaution $(\hat \mu - \hat \sigma)$ should be squared.

2)Cov. matrix $\Sigma(\mu, \sigma^2) $ shouldn't depend on $n$, it is asymptotic entity. So as $\Gamma$ which in this case is equal $\Gamma = \frac{3 \sigma^2}{2}$.

So the final answer is $T_n = \frac{2n^2 (\hat \mu - \hat \sigma)^2}{3 \hat \sigma^2}$


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