Standard Normal MGF and Even Moments

Worked Solution

How to Think About It: The MGF is the Swiss Army knife for computing moments -- the $n$-th moment is just the $n$-th derivative of the MGF evaluated at $t = 0$. For the standard normal, the MGF has a beautifully simple form ($e^{t^2/2}$), and differentiating it reveals a recursion: each even moment is $(2n-1)$ times the previous one. These are the "double factorial" moments. Once you know this pattern, you can rattle off $E[X^2] = 1$, $E[X^4] = 3$, $E[X^6] = 15$ without computation.

Quick Estimate: By symmetry, all odd moments are zero. The even moments grow as $(2n-1)!! = 1 \cdot 3 \cdot 5 \cdots (2n-1)$. So $E[X^2] = 1$, $E[X^4] = 3$, $E[X^6] = 15$, $E[X^8] = 105$, etc.

Formal Solution:

*Part (a): Deriving the MGF*

$$M_X(t) = E[e^{tX}] = \int_{-\infty}^{\infty} e^{tx} \cdot \frac{1}{\sqrt{2\pi}} e^{-x^2/2} \, dx = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-(x^2 - 2tx)/2} \, dx$$

Complete the square in the exponent: $$x^2 - 2tx = (x - t)^2 - t^2$$

So: $$M_X(t) = \frac{e^{t^2/2}}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-(x-t)^2/2} \, dx = e^{t^2/2}$$

since the remaining integral is the total probability of a $N(t, 1)$ density, which equals 1.

$$\boxed{M_X(t) = e^{t^2/2}}$$

*Deriving the recursion:*

The MGF can be expanded as a power series: $$M_X(t) = e^{t^2/2} = \sum_{n=0}^{\infty} \frac{(t^2/2)^n}{n!} = \sum_{n=0}^{\infty} \frac{t^{2n}}{2^n \cdot n!}$$

But also, by definition: $$M_X(t) = \sum_{k=0}^{\infty} \frac{E[X^k]}{k!} t^k$$

Comparing coefficients of $t^{2n}$: $$\frac{E[X^{2n}]}{(2n)!} = \frac{1}{2^n \cdot n!}$$

So: $$E[X^{2n}] = \frac{(2n)!}{2^n \cdot n!} = (2n-1)!!$$

where $(2n-1)!! = 1 \cdot 3 \cdot 5 \cdots (2n-1)$ is the double factorial.

The recursion follows immediately: $$E[X^{2n}] = (2n-1) \cdot E[X^{2n-2}]$$

Alternatively, differentiate the MGF: $M_X'(t) = t \cdot M_X(t)$, which gives the recursion $E[X^{2n}] = (2n-1) E[X^{2n-2}]$ by comparing Taylor coefficients.

*Part (b): Explicit moments*

Using the recursion with $E[X^0] = 1$: $$E[X^2] = 1 \cdot E[X^0] = 1$$ $$E[X^4] = 3 \cdot E[X^2] = 3$$ $$E[X^6] = 5 \cdot E[X^4] = 15$$

*Part (c): Sanity-checking simulated normals*

Given a sample $X_1, \ldots, X_N$ that should be i.i.d. $N(0,1)$, you can test normality by checking whether the sample moments match the theoretical values:

1. Variance check: The sample second moment $\hat{m}_2 = \frac{1}{N}\sum X_i^2$ should be close to 1. Under the null, $N \hat{m}_2 \sim \chi^2_N$, so a z-test on $\hat{m}_2$ works for large $N$.

1. Kurtosis check: The sample fourth moment ratio (excess kurtosis) is $\hat{\kappa} = \hat{m}_4 / \hat{m}_2^2 - 3$, which should be near 0 for a normal. Under the null, $\hat{\kappa}$ has asymptotic variance $24/N$, so $\hat{\kappa} / \sqrt{24/N}$ is approximately $N(0,1)$.

1. Sixth moment check: Similarly, $\hat{m}_6 / \hat{m}_2^3$ should be close to 15. This catches heavier-tailed deviations that kurtosis might miss.

1. Jarque-Bera test: Combine the skewness ($\hat{m}_3/\hat{m}_2^{3/2}$, should be 0) and kurtosis checks into a single chi-squared test with 2 degrees of freedom.

In practice, for RNG validation you would also check: uniformity of the CDF transform (Kolmogorov-Smirnov), serial correlation of the samples, and whether extreme quantiles appear at the right frequency.

Answer: (a) $M_X(t) = e^{t^2/2}$, with recursion $E[X^{2n}] = (2n-1) \cdot E[X^{2n-2}]$. (b) $E[X^2] = 1$, $E[X^4] = 3$, $E[X^6] = 15$. (c) Compare sample moments against theoretical values using z-tests (for individual moments) or the Jarque-Bera test (for skewness and kurtosis jointly), with asymptotic variance formulas derived from the moments themselves.