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The harmonic mean of positive numbers $x_1,\dots ,x_n$ is $H=n/({\sum }_{i}1/x_i)$. The Jensen inequality for $1/x$ gives $H\le \overline{x}$, with equality exactly when all entries coincide.

Write $S=R+\delta$ and let $C_i=\delta +{\sum }_{j\ne i}{b}_j$. Player $i$'s payoff in the proportional contest with outside effort is $u_i(b)=v_{i}b/(b+C_i)-b$. When $C_i>0$, this is strictly concave andThe best response is zero when $v_i\le C_i$, and otherwise is $\sqrt{v_i{C}_i}-C_i$. Thus the nonnegative-effort Karush-Kuhn-Tucker conditions at a pure strategy Nash equilibrium giveFor an active player the first-order equation is $v_i(S-b_i)=S^2$. For an inactive player the derivative at zero is $v_i/S-1$. Strict concavity makes these conditions sufficient as well as necessary.

If $\delta =0$, an equilibrium cannot have just one active player: that player would win with certainty and could lower its positive effort. An all-zero profile also cannot be an equilibrium under the usual completion of proportional allocation at zero: at least one player can gain by investing an arbitrarily small amount. Consequently there are at least two active players, so every $C_i$ is positive. If $\delta >0$ this positivity is automatic. The allocation rule's otherwise undefined all-zero value at $\delta =0$ is therefore immaterial to the equilibrium calculation.

Let $k=\hat{n}>0$, $H_k={\sum }_{i=1}^{k}1/v_i$, and use the harmonic mean ${\bar{v}}_k=k/H_k$. Summing the active efforts givesThe positive root supplies the total-effort formula for a proportional contest with outside effort:At $\delta =0$ this reduces to $R=(k-1){\bar{v}}_k/k$. There is also a genuine zero-active case: if $\delta \ge v_1$, then $k=0$ and $R=0$; no harmonic mean of an empty family is needed.

For a fully explicit active-set rule, setThe equilibrium equation is $G(S)=0$. For $\delta >0$, $G$ is strictly decreasing from $+\infty$ to $-1$. For $\delta =0$ its limit at zero is $n-1>0$, and it is strictly decreasing wherever a zero could occur. Hence the positive root is unique. Moreover $S\ge \delta$, since its equation gives $1-\delta /S\ge 0$, and $b_i=S(1-S/v_i{)}_{+}$ supplies an equilibrium with total effort $S-\delta$.

Player $j$ is active exactly when $v_j>S$, equivalently $G(v_j)<0$. Ordering the valuations, including ties, givesTherefore the active-set threshold for a proportional contest with outside effort isThe qualifying indices form a prefix because $G$ is decreasing. Equality excludes the marginal player, as required by strict positivity of effort. Equivalently, the positive root $S_k$ of the displayed quadratic must satisfy $v_k>S_k\ge v_{k+1}$, with $v_{n+1}=0$. These formulas include one active player when $\delta >0$ and exclude that case when $\delta =0$.

A stable multiplicative sensitivity factor $s_i$ at detector pixel $i$ changes both means to $N_i=s_i{N}_0$. The noise-free ratio is still one, so flat-field correction is not required merely to remove that fixed pattern from the mean ratio. This is the principal advantage of using two matched flats rather than the variance of a single image.

However, the photon shot noise is not cancelled. At high counts the conditional ratio variance is $2/(g{N}_i)$. Pooling detector pixels gives $E^2\simeq (2/g)⟨1/N_i⟩$, where brackets denote a spatial average. If the formula uses $\overline{N}=⟨N_i⟩$, it returnsThe inequality follows from the Jensen inequality for $1/x$ on positive signals. For small fractional response variation, the relative bias is approximately minus its squared coefficient of variation. A locally nearly uniform region or a fit using detector pixel-dependent variances avoids this bias from mixing the harmonic mean with the arithmetic mean. Stable response cancellation also presumes the conversion gain itself is uniform; gain variations, offset errors or changing sensitivity do not obey that simple cancellation. The fixed response pattern cancels in the ratio mean; unequal shot-noise levels can still bias a globally pooled gain estimate.

With $k>0$ active players of harmonic mean valuation ${\bar{v}}_k$, unit-cost Nash equilibrium effort satisfies $R+\delta =\frac{{\bar{v}}_k}{2k}[(k-1)+\sqrt{(k-1{)}^2+4k\delta /{\bar{v}}_k}]$. Sum the active-player first-order equations $b_i=(R+\delta )(1-(R+\delta )/v_i)$ to obtain the quadratic. If $\delta$ is at least the largest valuation, no one is active and $R=0$. With no outside effort, at least two players must be active.