Sunday 23 August 2026

Eigenvalues

Write $R=(1-\rho )I_n+\rho J$ with $J=\mathbf{1}{\mathbf{1}}^{\top }$ the all-ones matrix. The key observation — the step a candidate should reach for immediately — is that $J$ is rank one, so its spectrum is trivial: $J\mathbf{1}=n\mathbf{1}$, and $Jv=0$ for every $v⟂\mathbf{1}$. Hence $J$ has eigenvalue $n$ once (eigenvector $\mathbf{1}$) and eigenvalue $0$ with multiplicity $n-1$ (the orthogonal complement of $\mathbf{1}$).

Adding a multiple of the identity shifts every eigenvalue by the same amount and keeps the eigenvectors, so the eigenvalues of $R=(1-\rho )I_n+\rho J$ are

$${\lambda }_1=(1-\rho )+\rho n=1+(n-1)\rho \text{(multiplicity }1\text{)},$$

$${\lambda }_2=1-\rho \text{(multiplicity }n-1\text{)}.$$

As a check, the trace is ${\lambda }_1+(n-1){\lambda }_2=1+(n-1)\rho +(n-1)(1-\rho )=n$, matching $trR=n$.

Positive semidefiniteness

A symmetric matrix is positive semidefinite iff all its eigenvalues are nonnegative. So we need both

$$1-\rho \ge 0\text{and}1+(n-1)\rho \ge 0,$$

which give $\rho \le 1$ and $\rho \ge -\frac{1}{n-1}$. Therefore

$$\rho \in [-\frac{1}{n-1},1].$$

The tempting wrong answer is $\rho \in [-1,1]$, obtained by thinking of $\rho$ as an ordinary pairwise correlation. That is necessary but not sufficient: a single pair can be perfectly anticorrelated, but $n$ assets cannot all be that negatively correlated with one another simultaneously.

Portfolio-variance interpretation

For the equal-weight portfolio $w=\frac{1}{n}\mathbf{1}$,

$$w^{\top }Rw=\frac{1}{n^2}{\mathbf{1}}^{\top }R\mathbf{1}=\frac{1}{n^2}(n+n(n-1)\rho )=\frac{1+(n-1)\rho }{n}.$$

This is exactly ${\lambda }_1/n$, since $\mathbf{1}$ is the eigenvector for ${\lambda }_1$. Requiring this variance to be nonnegative recovers the binding constraint $\rho \ge -\frac{1}{n-1}$.

As $n\to \infty$, the lower bound $-\frac{1}{n-1}\to 0^{-}$. Interpretation: a large book of assets cannot maintain a common negative pairwise correlation. The most negative correlation compatible with all pairs being equal shrinks to zero, because diversification would otherwise drive the equal-weight portfolio variance below zero — impossible.

Final answers:

• Eigenvalues: $1+(n-1)\rho$ once, and $1-\rho$ with multiplicity $n-1$.
• $R$ is PSD iff $\rho \in [-\frac{1}{n-1},1]$.
• Equal-weight variance $={\displaystyle \frac{1+(n-1)\rho }{n}}$; its nonnegativity is the binding lower bound, which tends to $0$ as $n\to \infty$.

The weight and its two moments

With $p(x)=(2\pi {)}^{-1/2}{e}^{-x^2/2}$ and $q(x)=(2\pi {\sigma }^2{)}^{-1/2}{e}^{-x^2/(2{\sigma }^2)}$, $$w(x)=\frac{p(x)}{q(x)}=\sigma \exp (-\frac{x^2}{2}(1-\frac{1}{{\sigma }^2}\left)\right).$$

First moment. By construction the weight integrates the target against the proposal, so $${𝔼}_q[w]=\int \frac{p(x)}{q(x)}q(x)dx=\int p(x)dx=1.$$ This holds for every ${\sigma }^2$; it is what makes ordinary (non-normalized) importance sampling unbiased.

Second moment. This is the step that carries all the information: $${𝔼}_q[w^2]=\int \frac{p(x{)}^2}{q(x)}dx.$$ Now $$\frac{p(x{)}^2}{q(x)}=\frac{(2\pi {)}^{-1}{e}^{-x^2}}{(2\pi {\sigma }^2{)}^{-1/2}{e}^{-x^2/(2{\sigma }^2)}}=\frac{\sigma }{\sqrt{2\pi }}\exp (-(1-\frac{1}{2{\sigma }^2}\left)x^2\right).$$ The Gaussian integral $\int e^{-a{x}^2}dx=\sqrt{\pi /a}$ converges iff the coefficient $a=1-\frac{1}{2{\sigma }^2}>0$, i.e. iff ${\sigma }^2>\frac{1}{2}$. When it does, $${𝔼}_q[w^2]=\frac{\sigma }{\sqrt{2\pi }}\sqrt{\frac{\pi }{1-1/(2{\sigma }^2)}}=\frac{\sigma }{\sqrt{2-1/{\sigma }^2}}=\frac{{\sigma }^2}{\sqrt{2{\sigma }^2-1}}.$$ For ${\sigma }^2\le \frac{1}{2}$ the integral diverges: ${𝔼}_q[w^2]=+\infty$. (This is exactly $1+{\chi }^2(p\Vert q)$, the chi-square divergence, so the divergence has a name.)

Why $\rho$ is a ratio of these moments. By the law of large numbers, $\frac{1}{n}\sum w(X_i)\to {𝔼}_q[w]=1$ and $\frac{1}{n}\sum w(X_i{)}^2\to {𝔼}_q[w^2]$ (the latter finite only when ${\sigma }^2>\frac{1}{2}$). Hence $$\rho ({\sigma }^2)={lim}_{n\to \infty }\frac{\widehat{\mathrm{E}\mathrm{S}\mathrm{S}}}{n}=\frac{(\frac{1}{n}\sum w{)}^2}{\frac{1}{n}\sum w^2}=\frac{{𝔼}_q[w{]}^2}{{𝔼}_q[w^2]}=\frac{1}{1+{\mathrm{V}\mathrm{a}\mathrm{r}}_q(w)/{𝔼}_q[w{]}^2}.$$ When ${𝔼}_q[w^2]=\infty$ the denominator diverges and $\rho =0$: the effective sample size is $o(n)$.

The closed form and its optimum

For ${\sigma }^2>\frac{1}{2}$, $$\rho ({\sigma }^2)=\frac{1}{{𝔼}_q[w^2]}=\frac{\sqrt{2{\sigma }^2-1}}{{\sigma }^2},$$ and $\rho ({\sigma }^2)=0$ for ${\sigma }^2\le \frac{1}{2}$.

Write $t={\sigma }^2$ and maximize $g(t)=t^{-1}\sqrt{2t-1}$: $$g^{\prime }(t)=\frac{t(2t-1{)}^{-1/2}-(2t-1{)}^{1/2}}{t^2}=0⟺t=2t-1⟺t=1.$$ So the maximum is at ${\sigma }^2=1$, where $\rho (1)=\sqrt{1}/1=1$. This is the sanity check: $q=p$ gives constant weights and full effective sample size, $\rho =1$.

Solving $\rho =\frac{1}{2}$

Set $\sqrt{2t-1}/t=\frac{1}{2}$ with $t>\frac{1}{2}$. Squaring, $$\frac{2t-1}{t^2}=\frac{1}{4}⟹t^2-8t+4=0⟹t=4\pm 2\sqrt{3}.$$ Both roots exceed $\frac{1}{2}$ (indeed $4-2\sqrt{3}=(\sqrt{3}-1{)}^2\approx 0.536$ and $4+2\sqrt{3}\approx 7.46$), so both are admissible: $$\boxed{{\sigma }^2=4-2\sqrt{3}\approx 0.536\text{and}{\sigma }^2=4+2\sqrt{3}\approx 7.46.}$$

Interpretation. The efficiency curve $\rho$ is single-peaked at ${\sigma }^2=1$ and falls off on both sides, so a given efficiency level is hit once on each side. But the two sides are not symmetric in their failure mode. Making the proposal too narrow (${\sigma }^2\le \frac{1}{2}$) sends $\rho$ to a hard zero — the weight variance is literally infinite, the LLN for $\widehat{\mathrm{E}\mathrm{S}\mathrm{S}}$ fails, and a handful of samples in the target's tail carry all the mass. Making it too wide only degrades $\rho$ smoothly toward $0$ as ${\sigma }^2\to \infty$ (since $\rho ~\sqrt{2}{\sigma }^{-1}$), with all moments finite throughout. The lesson every practitioner eventually relearns: overdisperse the proposal, never underdisperse it.

Closing note

The tempting error is to treat $\rho ({\sigma }^2)$ as finite and positive for all ${\sigma }^2$ — the closed form $\sqrt{2{\sigma }^2-1}/{\sigma }^2$ formally returns a complex number for ${\sigma }^2<\frac{1}{2}$, which quietly hides the real event: the second moment has diverged and the estimator's asymptotic variance is infinite. The convergence condition $a=1-1/(2{\sigma }^2)>0$ is the whole point of the problem, not a technicality.