Recurrence for Standard Normal Moments

Why Standard Normal Moments Have a Simple Pattern

The standard normal distribution appears throughout statistical theory, so calculations involving its moments recur in many settings. Rather than integrating separately for every power, we can use a single recurrence to obtain each higher even moment from the moment two orders below it. Symmetry supplies the odd moments, and together these facts give all moments through any fixed integer order.

Let \(Z\sim N(0,1)\). Its density is \(\phi(x)=(2\pi)^{-1/2}\exp(-x^2/2)\), for \(x\in\mathbb{R}\). The density is symmetric about zero, and its derivative satisfies \(\phi'(x)=-x\phi(x)\). These two properties drive the calculations below.

All the moments in this definition exist as finite expectations. Indeed, for each fixed nonnegative integer \(r\), the absolute-moment integral is proportional to \(\int_{-\infty}^{\infty}|x|^r e^{-x^2/2}\,dx\), which is finite: for sufficiently large \(|x|\), the polynomial factor is bounded by a constant times \(e^{x^2/4}\), leaving an integrable tail bounded by a constant times \(e^{-x^2/4}\). Thus the signed integrals used below converge absolutely. Also, for every fixed nonnegative integer \(q\), \(|x|^q\phi(x)\to0\) as \(|x|\to\infty\), because \(q\log|x|-x^2/2\to-\infty\). This will make the boundary term in integration by parts vanish.

Odd Moments Vanish by Symmetry

The density has the same value at \(x\) and \(-x\). When \(r\) is odd, the factor \(x^r\) changes sign under reflection, so contributions from opposite sides of zero cancel. Absolute integrability, established above, justifies the change of variable over the whole real line.

To see this explicitly, substitute \(y=-x\) in the moment integral. Since \(\phi(-y)=\phi(y)\) and \((-y)^r=-y^r\) for odd \(r\),

The moment is finite, so \(m_r=-m_r\) implies \(m_r=0\). In particular, \(E[Z]=0\), and every higher odd signed moment is zero as well.

Deriving the Moment Recurrence

For even moments, integration by parts reduces the order by two. The derivative identity \(\phi'(x)=-x\phi(x)\) lets us replace one factor of \(x\) in the integrand with a derivative of the density.

Proof. Fix an integer \(r\geq2\). Apply integration by parts on the finite interval \([-A,A]\), using \(u=x^{r-1}\) and \(dv=x\phi(x)\,dx=-\phi'(x)\,dx\). Then \(du=(r-1)x^{r-2}\,dx\) and \(v=-\phi(x)\), giving

As \(A\to\infty\), the boundary term tends to zero: both \(A^{r-1}\phi(A)\) and \((-A)^{r-1}\phi(-A)\) tend to zero by the Gaussian exponential decay noted above. Both integrals converge to their corresponding whole-line integrals, since their integrands are absolutely integrable. Taking the limit therefore yields

This proves the recurrence for every integer \(r\geq2\). \(\square\)

The recurrence applies to odd orders too, but it does not replace the symmetry argument: starting from \(m_1=0\), it gives \(m_3=2m_1=0\), \(m_5=4m_3=0\), and so on. For even orders, it starts from \(m_0=1\) and gives nonzero values.

Even Moments Through Order Eight

Applying the recurrence from \(m_0=1\) gives the following values. The odd-order entries follow from symmetry, not from treating the recurrence as a formula that begins at \(m_0\).

In compact form, the even moments through order eight are \(E[Z^2]=1\), \(E[Z^4]=3\), \(E[Z^6]=15\), and \(E[Z^8]=105\). Each is the product of the positive odd integers below its order. The recurrence also provides a direct check: multiplying each even moment by the next odd factor gives the next even moment.

Common Mistake: Confusing Signed and Absolute Moments

A frequent error is to conclude that an odd power contributes nothing in every calculation because its expectation is zero. That conclusion holds for the signed moment \(E[Z^r]\) when \(r\) is odd; it does not hold for \(E[|Z|^r]\), where the absolute value removes the sign cancellation.

For instance, \(E[Z^3]=0\), but \(E[|Z|^3]>0\). The integrand \(|x|^3\phi(x)\) is nonnegative and is positive except at \(x=0\), so its integral cannot vanish. The correction is to inspect the quantity actually being averaged: symmetry cancels positive and negative contributions only when the integrand retains the odd sign.

Another tempting shortcut is to use \(m_r=(r-1)m_{r-2}\) for \(r=1\). The integration-by-parts derivation requires \(r\geq2\); moreover, the recurrence at \(r=1\) would refer to \(m_{-1}\), which is outside the defined sequence. Obtain \(m_1=0\) directly from symmetry, and use the recurrence only at integer orders \(r\geq2\).

Why the Recurrence Is Useful

The recurrence replaces a sequence of separate integrals with a short calculation, and symmetry eliminates every odd signed moment without integration. Together they provide a reliable way to evaluate polynomial expectations under the standard normal distribution, check variance calculations, and recognize the fourth moment \(3\) that appears in standardized moment summaries. The key is to track the moment order carefully and distinguish signed powers from absolute powers.