Let (\mathbb {R}, \{ N(\mu , \sigma ^2)\} _{\mu \in \mathbb {R}, \sigma > 0}) be the statistical model of a normal random variable X. Let
m_ k(\mu , \sigma ) = \mathbb {E}[X^ k]
denote the k-th moment of X. Let \psi : \mathbb {R} \times (0, \infty ) \to \mathbb {R}^2 be defined by \psi (\mu , \sigma ) = (m_1(\mu , \sigma ), m_2(\mu , \sigma )). (Since we have two parameters of interest,\mu and \sigma, it makes sense to work with the first two moments. The hope is that the two moments will uniquely determine the paramters of interest \mu and \sigma.)
Express m_1(\mu , \sigma ) and m_2(\mu , \sigma ) in terms of \mu and \sigma.
We have that X follows a normal distribution with mean \mu and variance \sigma^2. Therefore, the k-th moment of X can be found using the following integral:
m_k(\mu,\sigma) = \int_{-\infty}^{\infty} x^k \frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{(x-\mu)^2}{2\sigma^2}} dx
Let's find the first moment:
m_1(\mu,\sigma) = \int_{-\infty}^{\infty} x \frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{(x-\mu)^2}{2\sigma^2}} dx
We recognize this as the pdf of a normal distribution with mean \mu and variance \sigma^2. Therefore, it integrates to 1 and we can simplify the expression to:
m_1(\mu,\sigma) = \mu
Now, let's find the second moment:
m_2(\mu,\sigma) = \int_{-\infty}^{\infty} x^2 \frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{(x-\mu)^2}{2\sigma^2}} dx
We can simplify this expression by using the identity that the variance of a normal distribution is \sigma^2, which is equal to m_2(\mu,\sigma)-m_1(\mu,\sigma)^2. Therefore, we have:
m_2(\mu,\sigma) = \sigma^2 + \mu^2
So, in terms of \mu and \sigma, we have:
m_1(\mu,\sigma) = \mu
m_2(\mu,\sigma) = \sigma^2 + \mu^2
The moment generating function of a normal random variable with mean \(\mu\) and variance \(\sigma^2\) is given by:
\[M_X(t) = e^{\mu t+\frac{1}{2}\sigma ^{2}t^{2}}\]
The \(k\)-th moment of a random variable can be obtained by taking the \(k\)-th derivative of the moment generating function and evaluating it at \(t = 0\):
\[m_k(\mu , \sigma ) = \frac{{d^k}}{{dt^k}}\left(M_X(t)\right) \bigg|_{t=0}\]
Let's calculate the first and second moments, \(m_1(\mu , \sigma )\) and \(m_2(\mu , \sigma )\):
First moment:
\[m_1(\mu , \sigma ) = \frac{d}{dt}\left(M_X(t)\right) \bigg|_{t=0}\]
\[ = \frac{d}{dt}\left( e^{\mu t+\frac{1}{2}\sigma ^{2}t^{2}}\right) \bigg|_{t=0}\]
Using the chain rule, we have:
\[ = \left(\mu + \sigma^2 t\right) e^{\mu t+\frac{1}{2}\sigma ^{2}t^{2}} \bigg|_{t=0}\]
\[ = \mu\]
Therefore, the first moment \(m_1(\mu , \sigma )\) is equal to \(\mu\).
Second moment:
\[m_2(\mu , \sigma ) = \frac{d^2}{dt^2}\left(M_X(t)\right) \bigg|_{t=0}\]
\[ = \frac{d^2}{dt^2}\left( e^{\mu t+\frac{1}{2}\sigma ^{2}t^{2}}\right) \bigg|_{t=0}\]
Using the chain rule again, we have:
\[ = \left(\mu + \sigma^2 t\right)^2 e^{\mu t+\frac{1}{2}\sigma ^{2}t^{2}} \bigg|_{t=0}\]
\[ = \mu^2 + \sigma^2\]
Therefore, the second moment \(m_2(\mu , \sigma )\) is equal to \(\mu^2 + \sigma^2\).
Hence, we have expressed the first moment \(m_1(\mu , \sigma )\) as \(\mu\) and the second moment \(m_2(\mu , \sigma )\) as \(\mu^2 + \sigma^2\).