Optimization in One Variable

1Introduction

Optimization chooses the best value of a decision variable: how many lots to buy, which leverage to hold, what yield prices a bond. In one variable the tools are elementary: a zero derivative at the top, the second derivative to tell tops from bottoms, and the endpoints. A sound answer settles three questions: is the critical point the global top, does a constraint bind, and how much does the exact location matter? This module settles all three for the sizing problem every leveraged trader faces.

Sections 2–4 build the tools: first- and second-order conditions, concavity and global optima, and constraints that create corner solutions. Section 5 derives the binary Kelly bet exactly. Section 6 solves equations with no closed form — the empirical Kelly condition and a bond's yield — by bisection and Newton's method. Section 7 measures the sensitivity of an optimum, introduces the envelope theorem and resolves the case; Section 8 does it all in Python.

2First- and Second-Order Conditions

Optimization starts with a function of one decision variable, such as growth against leverage, and asks where it is largest. Calculus answers in two steps. The first-order condition locates the candidates; the second-order condition sorts them.

Local and global extrema

Every global maximum is a local one, but not conversely. A trader cares about the global maximum, while calculus sees only local information: the slope and curvature at a point. Section 3 shows when local information is enough.

The first-order condition

Proof. Let $x^{\ast }$ be a local maximum. For small $\Delta >0$, $h(x^{\ast }+\Delta )-h(x^{\ast })\le 0$, so the right difference quotient $[h(x^{\ast }+\Delta )-h(x^{\ast })]/\Delta$ is $\le 0$ and its limit gives $h^{\prime }(x^{\ast })\le 0$. For small $\Delta <0$ the numerator is still $\le 0$ but the denominator is negative, so the left quotient is $\ge 0$ and $h^{\prime }(x^{\ast })\ge 0$. Hence $h^{\prime }(x^{\ast })=0$; a minimum reverses the signs. $■$

Three warnings follow from the proof. The condition is necessary, not sufficient: $h(x)=x^3$ has $h^{\prime }(0)=0$ and no extremum. It applies only at interior points; at an endpoint the one-sided argument yields an inequality instead (Section 4). And a kink, such as $|x|$ at 0, is a candidate even though no derivative exists there.

The second-order condition

Proof. Taylor's theorem with the Lagrange remainder (LM3, eq. (3.5)) gives $h(x^{\ast }+\Delta )=h(x^{\ast })+h^{\prime }(x^{\ast })\Delta +\frac{1}{2}{h}^{\prime \prime }(\xi ){\Delta }^2$ for some $\xi$ between $x^{\ast }$ and $x^{\ast }+\Delta$. The linear term vanishes by the first-order condition. If $h^{\prime \prime }(x^{\ast })<0$, continuity keeps $h^{\prime \prime }<0$ on a small interval around $x^{\ast }$, so $h(x^{\ast }+\Delta )<h(x^{\ast })$ for every small $\Delta \ne 0$. The case $h^{\prime \prime }(x^{\ast })>0$ is symmetric. $■$

When $h^{\prime \prime }(x^{\ast })=0$, look at higher derivatives. If the first non-zero derivative at $x^{\ast }$ has order $k$, a Taylor expansion to order $k$ fixes the sign of $h(x^{\ast }+\Delta )-h(x^{\ast })$. For even $k$ the point is an extremum, a maximum if $h^{(k)}(x^{\ast })<0$; for odd $k$ the graph crosses its tangent, with no extremum. Exhibit 2 collects the cases.

Growth as a function of leverage

Now the decision this module is about. A position of $f$ units of an asset per unit of equity turns a return $R$ into $fR$; one period multiplies wealth by $1+fR$. Wealth compounds, so the natural objective is the expected log growth per period,

Section 5 shows why maximizing $g$ maximizes long-run wealth; here only its shape matters. Expand $\ln (1+x)\approx x-x^2/2$ (LM3, eq. (3.3)) with $x=fR$, take expectations, and use $\mathbb{E}[R^2]={\sigma }^2+{\mu }^2$ (the population form of LM3's identity (3.7)):

The neglected terms are of third order in $fR$. This quadratic model is a downward parabola. Its first-order condition $\mu -f({\sigma }^2+{\mu }^2)=0$ gives

and $g^{\prime \prime }=-({\sigma }^2+{\mu }^2)<0$ confirms a strict maximum. The second step drops ${\mu }^2$, which is smaller than ${\sigma }^2$ by the factor $(\mu /\sigma {)}^2$; for 0050's daily returns $(\hat{\mu }/s{)}^2=0.0049$. The Sharpe ratio $\mathrm{SR}:=(\mu -r_f)/\sigma$ is the mean return in excess of a risk-free rate $r_f$, per unit of standard deviation (LM38 studies its estimation). With $r_f=0$, as here, it is $\mu /\sigma$ per period.

At the optimum $g(f_q)\approx {\mu }^2/(2({\sigma }^2+{\mu }^2))\approx \frac{1}{2}{\mathrm{SR}}^2$: maximal growth is about half the squared Sharpe ratio per period. There the drag term $\frac{1}{2}{f}_q^2({\sigma }^2+{\mu }^2)$ is exactly half the levered mean $f_q\mu$, so at the optimum, volatility drag consumes half the levered arithmetic return. On 0050 at the exact optimum, a levered arithmetic mean of 114.83% a year becomes log growth of 58.84%: 51% is kept. LM42 derives the continuous-time versions, $f^{\ast }=(\mu -r_f)/{\sigma }^2$ and $g^{\ast }=r_f+{\mathrm{SR}}^2/2$.

3Concavity and Global Optima

The second-derivative test is local: it compares $x^{\ast }$ only with its neighbours. A sizing decision needs a global answer — no leverage anywhere does better. Concavity converts local information into that global guarantee.

For twice-differentiable functions there is a working test, stated here without proof (Simon and Blume 1994, chapter 21). On an open interval, $h$ is concave exactly when $h^{\prime \prime }\le 0$ throughout, and $h^{\prime \prime }<0$ implies strict concavity. The converse of the last statement fails: $-x^4$ is strictly concave although its second derivative vanishes at 0.

Proof (when $h^{\prime \prime }\le 0$ on $I$). For any $x\in I$, Taylor's theorem (3.5) gives $h(x)=h(x^{\ast })+h^{\prime }(x^{\ast })(x-x^{\ast })+\frac{1}{2}{h}^{\prime \prime }(\xi )(x-x^{\ast }{)}^2\le h(x^{\ast })$, because $h^{\prime }(x^{\ast })=0$ and $h^{\prime \prime }(\xi )\le 0$. If $h^{\prime \prime }<0$ on $I$, the inequality is strict for $x\ne x^{\ast }$. $■$ The Deep Dive proves the theorem without second derivatives.

One corollary is used constantly. A differentiable, strictly concave function has at most one critical point, and if it has one, that point is the global maximizer. Finding the optimum then becomes a pure root-finding problem, the subject of Section 6.

The log-growth function is strictly concave

Apply this to the sample version of (4.1). For returns $R_1,\dots ,R_T$, the empirical growth function is

defined on the set $\mathcal{D}$ of leverages with $1+f{R}_t>0$ for every $t$. If the sample contains both gains and losses, $\mathcal{D}$ is the open interval $(-1/R_{\max },1/|R_{\min }|)$. Differentiating term by term,

Uniqueness follows from (4.6), since ${\bar{g}}^{\prime }$ is continuous and strictly decreasing on $\mathcal{D}$. For existence, let $f$ rise towards $1/|R_{\min }|$: the worst day's term $R_{\min }/(1+f{R}_{\min })$ tends to $-\infty$, so ${\bar{g}}^{\prime }\to -\infty$. As $f$ falls towards $-1/R_{\max }$, the best day's term tends to $+\infty$, and so does ${\bar{g}}^{\prime }$.

The intermediate value theorem says that a continuous function that is positive at one point and negative at another is zero somewhere between them. It is stated here without proof (Stewart, Clegg and Watson 2021, chapter 2). So ${\bar{g}}^{\prime }$ crosses zero, and exactly once. The same edges show that $\bar{g}(f)\to -\infty$: near $1/|R_{\min }|$ the worst day all but wipes out the account.

For 0050 the largest daily loss and gain were both exactly 10%: −10.00% on 7 April 2025 and +10.00% on 31 July 2026. Ten per cent is Taiwan's daily price limit, in force since 1 June 2015 (it was 7% before). So $\mathcal{D}=(-10,10)$, and at 10× a single limit-down day ends the experiment. Exhibit 3 plots $\bar{g}$ for 0050 against the quadratic model.

4Constraints and Corner Solutions

Real positions face limits. A cash account cannot borrow, many mandates cannot sell short, and brokers cap margin. Each limit confines $f$ to an interval, and the optimum may then sit on the boundary, where the derivative need not be zero.

The theorem is stated without proof (Stewart, Clegg and Watson 2021, chapter 4). It turns optimization on $[\alpha ,\beta ]$ into a finite search, the closed-interval method. Evaluate $h$ at its critical points in $(\alpha ,\beta )$ and at both endpoints, and take the largest value; the method needs continuity only, not concavity.

First-order conditions at a corner

At an endpoint the argument of Fermat's theorem runs on one side only, so it delivers an inequality. If a differentiable $h$ attains its maximum on $[\alpha ,\beta ]$ at $\beta$, every left difference quotient $[h(\beta -\Delta )-h(\beta )]/(-\Delta )$ is $\ge 0$, so $h^{\prime }(\beta )\ge 0$. Likewise a maximum at $\alpha$ requires $h^{\prime }(\alpha )\le 0$. Collecting the three cases,

These are the one-variable Karush–Kuhn–Tucker (KKT) conditions, which LM7 extends with Lagrange multipliers. For a concave $h$ they are also sufficient, by the tangent-line inequality of the Deep Dive in Section 3.

The derivative at a binding corner has an economic reading. If the optimum is $\beta$ and $h^{\prime }(\beta )>0$, relaxing the limit to $\beta +\Delta$ raises the objective by about $h^{\prime }(\beta )\Delta$ when $\Delta$ is small. That derivative is the shadow price of the constraint: the rate at which relaxing it raises the objective, a marginal value rather than a total. In LM7 it becomes the Lagrange multiplier.

Proof. If $x^{\ast }\in [\alpha ,\beta ]$ there is nothing to show. If $x^{\ast }>\beta$, then $h^{\prime }$ is non-increasing with $h^{\prime }(x^{\ast })=0$, so $h^{\prime }\ge 0$ on $[\alpha ,\beta ]$ and $h$ is non-decreasing there. Hence $\beta$ is a maximizer, and the only one if $h$ is strictly concave, since then $h^{\prime }>0$ on $[\alpha ,\beta ]$. The case $x^{\ast }<\alpha$ is symmetric. $■$

Two corners in the 0050 data

A long-only investor without a margin account can still reach any leverage between 0 and 2. Mixing cash, 0050 and the daily 2× fund 00631L (LM3) does it, if rebalanced daily and ignoring costs and replication basis. The feasible set is $[0,2]$. Exhibit 4 shows what in-sample optimization recommends year by year, with and without that interval.

Two facts stand out. Year by year the unconstrained optimum ranges from a 4.46× short position to 19.73× long. And every constrained optimum is a corner: 0 in the three years with $\hat{\mu }<0$, 2 in the eight with $f^{\ast }>2$.

5The Binary Kelly Bet

The simplest sizing problem has one source of risk, two outcomes and an exact solution. It exposes the confusion between position size and risk size, and shows how sharply the optimum depends on the win probability. LM41 extends it to general discrete bets.

Why maximize expected log growth

A trade gains a fraction $b$ of the position with probability $p$ and loses a fraction $a$ with probability $q=1-p$, independently of other trades. With a position of $f$ times equity, one trade multiplies wealth by $1+fb$ or $1-fa$. After $N$ trades with $j$ wins,

By the law of large numbers (LM17), $j/N\to p$, so the per-trade log growth converges to

If $g(f)>g(\tilde{f})$, the ratio of the two wealth paths behaves like $e^{N[g(f)-g(\tilde{f})]}$ and grows without bound. The fraction with the highest $g$ therefore ends up with more wealth than any other fixed fraction, with probability one (Breiman 1961; Thorp 2006). That is Kelly's (1956) criterion: maximize the expected log growth. MacLean, Thorp and Ziemba (2011) collect the classic papers on it.

Deriving the Kelly position

From (4.7),

The growth function is strictly concave (the Knowledge Check of Section 3), so the root of $g^{\prime }$ is the unique global maximum. Setting $g^{\prime }(f)=0$ and cross-multiplying gives $pb(1-fa)=qa(1+fb)$, so $pb-qa=fab(p+q)=fab$, and

No approximation was made: (4.8) is exact. Substituting it back gives $1+f^{\ast }b=p(a+b)/a$ and $1-f^{\ast }a=q(a+b)/b$, so the maximal growth rate is

A rewriting makes (4.8) easier to read. Let $\pi :=a/(a+b)$ be the break-even probability, at which the expected return $pb-qa$ is zero. Then

The Kelly position is proportional to the probability edge $p-\pi$, with slope $1/a+1/b$. It is positive exactly when the expected return is positive. With no edge, $f^{\ast }=0$ is an interior optimum with $g^{\prime }(0)=0$; with a negative edge, a long-only rule gives the corner solution $f=0$.

Position size versus risk size

On a losing trade the account loses $fa$ of equity, the risk fraction. Multiplying (4.8) by $a$,

When the whole stake can be lost, $a=1$ and position equals risk. Then (4.11) is the classic Kelly formula $p-q/b$ for a bet paying $b$ to 1, which Thorp (2006) writes as $(bp-q)/b$. A stop-loss trade loses only $a$ of the position, so the two numbers differ by the factor $1/a$: 100 for a 1% stop.

Equation (4.11) carries a second message. Scaling $a$ and $b$ by the same factor leaves the payoff ratio $b/a$, $f^{\ast }a$, $1+f^{\ast }b$ and $1-f^{\ast }a$ unchanged, so by (4.9) it leaves $g^{\ast }$ unchanged. Only the payoff ratio and the win probability determine the attainable growth. The stop distance sets the position size that delivers it.

Sensitivity to the win probability

Equation (4.10) makes the sensitivity explicit: $d{f}^{\ast }/dp=1/a+1/b$, which is 150 for the +2%/−1% bet. One percentage point of win probability moves the Kelly position by 1.5 times equity. Exhibit 5 follows the bet as its true win probability falls, and asks what happens to a trader who keeps the 25× position.

Growth at a fixed 25× stays positive until $p$ falls to 0.415. There the true Kelly position is 12.26, so the trader holds about twice Kelly, the zero-growth point of Section 7. Below it the trader loses money in the long run even though every trade still has a positive expected return. At $p=0.40$ the expected return is $0.40\times 2\%-0.60\times 1\%=+0.2\%$ per unit of position, yet the 25× position shrinks wealth by 1.04% per trade.

6Root Finding: Bisection and Newton's Method

The binary bet is the exception: most first-order conditions have no closed form. The empirical Kelly condition ${\bar{g}}^{\prime }(f)=0$ is the sample version of $\mathbb{E}[R/(1+fR)]=0$. Once denominators are cleared, for 0050 it becomes a polynomial equation of degree 2,709, one less than the number of distinct non-zero daily returns.

A bond's yield and a project's internal rate of return are roots of polynomials too. Numerical root finding solves $u(x)=0$ to any accuracy, and for a smooth concave objective $h$, maximizing $h$ is root finding on $u=h^{\prime }$.

Bisection

By the intermediate value theorem (Section 3) every interval in the sequence contains a root. After $n$ steps the bracket has width $(x_{\mathrm{hi}}-x_{\mathrm{lo}})/2^n$, so its midpoint is within $(x_{\mathrm{hi}}-x_{\mathrm{lo}})/2^{n+1}$ of a root. A bracket of width $\epsilon$ takes

steps: one binary digit per step, or about 3.3 steps per decimal digit. The error bound shrinks by a constant factor each step, which is called linear convergence. Bisection needs only continuity and a sign change, and it cannot fail.

Newton's method

Newton's method replaces $u$ by its tangent at the current point, the first-order Taylor polynomial of LM3, and jumps to the tangent's root:

Applied to $u=h^{\prime }$ for a maximization, the step is $x_{n+1}=x_n-h^{\prime }(x_n)/h^{\prime \prime }(x_n)$: each iteration maximizes the second-order Taylor model of $h$ at $x_n$.

The speed comes from the remainder. Let $x^{\ast }$ be a root with $u^{\prime }(x^{\ast })\ne 0$ and write $e_n=x_n-x^{\ast }$. Taylor's theorem gives $0=u(x^{\ast })=u(x_n)+u^{\prime }(x_n)(x^{\ast }-x_n)+\frac{1}{2}{u}^{\prime \prime }({\xi }_n)(x^{\ast }-x_n{)}^2$ for some ${\xi }_n$ between $x_n$ and $x^{\ast }$. Dividing by $u^{\prime }(x_n)$ and using (4.12),

The error is squared at each step: quadratic convergence. Once $|e_n|$ is small, the number of correct digits roughly doubles per iteration. The constant $C=|u^{\prime \prime }(x^{\ast })/(2u^{\prime }(x^{\ast }))|$ says how small is small: squaring helps once $C|e_n|<1$.

The theorem is stated without its full proof; (4.13) is the core, and Burden, Faires and Burden (2016, chapter 2) supply the details. It is a local result. Far from the root the tangent can point anywhere. Where $u^{\prime }(x_n)\approx 0$ the step is huge, and a step can leave the domain or land near a different root.

Safeguarded methods combine the two ideas. They keep a bracket, accept a fast step (Newton or secant) when it lands inside it, and bisect otherwise. Brent's (1973) method, available as scipy.optimize.brentq, is the standard version: it never loses the bracket and converges superlinearly near the root.

The empirical Kelly condition

Newton's method on $u={\bar{g}}^{\prime }$ uses (4.5) and (4.6). Start from $f_0=0$, where ${\bar{g}}^{\prime }(0)=\hat{\mu }$ and ${\bar{g}}^{\prime \prime }(0)=-m_2$:

The quadratic rule (4.3) is exactly one Newton step from zero. Each later step refits the quadratic model at the current point. Exhibit 6 runs both methods on 0050. Bisection starts from the bracket $[0,9.9]$, since ${\bar{g}}^{\prime }(0)=\hat{\mu }>0$ and ${\bar{g}}^{\prime }(9.9)<0$: ${\bar{g}}^{\prime }$ falls towards $-\infty$ as $f$ approaches 10, where it is undefined.

7Sensitivity of the Optimum: Flat Tops and the Envelope Theorem

Two questions remain once an optimum is found: what does missing it cost, and how do the optimum and its value move with the inputs? The second-order term of a Taylor expansion answers the first, the envelope theorem the second. Together they resolve the case.

Flat tops

At an interior maximum the first-order term of the Taylor expansion vanishes, so

where the approximation drops terms of third order in $f-f^{\ast }$. The loss from mis-sizing is of second order in the sizing error. An error twice as large costs four times as much, and a small error costs very little. The top of a smooth maximum is flat.

In the quadratic model the flat top has an exact form. Write (4.2) as $g(f)=f\mu -\frac{1}{2}{f}^2{m}_2$, with $f^{\ast }=\mu /m_2$ and $g^{\ast }={\mu }^2/(2m_2)$. Substituting $f=c{f}^{\ast }$,

This is Thorp's (2006) $c(2-c)$. Exhibit 7 compares (4.15) with the exact curves for 0050 and for the +2%/−1% bet.