Unit 11, lesson 86

Liquidity risk and closeout horizons

How a liquidation path turns a position into spread cost, liquidation impact cost and inventory variance, how the liquidation horizon balances them, and what a liquidity adjustment to a mark or a risk figure measures.

  • Video 86
  • Stress testing and liquidity risk

A fund holds 100000100000 shares marked at the mid price and has to sell all of them. Any sale pays the spread, a fast sale also accepts worse prices as its own orders push the market, and a slow sale carries the unsold shares through whatever the price does in the meantime. How much of the mark should the fund expect to give up, and over how many days should it sell? This lesson models liquidation as a trading path, computes the expected spread and impact cost and the variance from carrying inventory, finds the horizon that balances them under a mean-variance criterion, and explains what such a liquidity adjustment does and does not measure.

Key terms

Liquidity risk

Definition Market liquidity risk is the risk that closing out a position realizes less than its mark, because the trades pay the bid-ask spread, move the execution price, and take time during which the price can change. It is distinct from funding liquidity risk, the risk of being unable to meet payments when they fall due, which is not treated here. In this lesson time tt is measured in trading days, and a long position of q>0q > 0 shares marked at the mid price S0S_0 is sold over [0,h][0, h] along a deterministic schedule. The inventory x(t)x(t) has x(0)=qx(0) = q and x(h)=0x(h) = 0, and the selling rate is v(t)=−x′(t)≥0v(t) = -x'(t) \ge 0 shares per day. The model assumes that the unaffected mid price is St=S0+σWtS_t = S_0 + \sigma W_t, with WW a standard Brownian motion under P\mathbb{P}, zero drift and σ>0\sigma > 0 in currency units per share per square root of a day, and that a share sold at time tt fetches St−b−ηv(t)S_t - b - \eta v(t) for constants b≥0b \ge 0 and η>0\eta > 0. The liquidation cost, measured against the mark, is

C=qS0−∫0h(St−b−ηv(t)) v(t) dt.\begin{aligned} C &= q S_0 - \int_0^h \bigl(S_t - b \\[0.4em] &\qquad - \eta v(t)\bigr)\,v(t)\,dt . \end{aligned}

Since xx is deterministic with x(h)=0x(h) = 0 and W0=0W_0 = 0, the product rule gives ∫0hv(t)Wt dt=∫0hx(t) dWt\int_0^h v(t) W_t\,dt = \int_0^h x(t)\,dW_t, and so

C=bq+η∫0hv(t)2 dt−σ∫0hx(t) dWt.\begin{aligned} C &= bq + \eta \int_0^h v(t)^2\,dt \\[0.4em] &\quad - \sigma \int_0^h x(t)\,dW_t . \end{aligned}

The cost is positive when the sale fetches less than the mark, and a rising price lowers it.

In plain words The mark is a reference price, not a promise of what the shares will fetch. Selling takes time, and each stretch of the schedule costs something, either through the trades themselves or through the price risk of the shares still held.

Example For a constructed position of q=100000q = 100000 shares with b=0.005b = 0.005, η=10−7\eta = 10^{-7} and σ=0.3\sigma = 0.3, sold at a constant rate over one day, the cost turns out to have mean 15001500 currency units and standard deviation about 1732117321 currency units, as the next cards show. The expected cost is small next to its uncertainty, so a figure for liquidity risk has to say which of the two it measures.

Spread cost

Definition The half-spread b≥0b \ge 0, in currency units per share, is the distance from the mid price to the bid, where a seller trades, or to the ask, where a buyer trades. Selling qq shares at the bid instead of the mid costs bqbq relative to a mid mark, whatever the schedule, because ∫0hv(t) dt=q\int_0^h v(t)\,dt = q. In this model bb is constant, so the spread cost does not depend on the horizon, and every schedule has E[C]≥bq\mathbb{E}[C] \ge bq. If the full spread is quoted as a fraction ss of the mid, then b=sS0/2b = sS_0/2. A short position bought back pays the same amount at the ask. A spread that widens in a stress is a separate scenario input, not something the model produces.

In plain words Crossing the spread costs half the spread on every share, however patiently the position is sold. It sets a floor under the expected cost.

Example With b=0.005b = 0.005 and q=100000q = 100000 the spread cost is 0.005⋅100000=5000.005 \cdot 100000 = 500 currency units. If the half-spread widened to 0.010.01, the spread cost alone would double to 10001000.

Liquidation impact cost

Definition Under linear temporary impact, a share sold at time tt fetches ηv(t)\eta v(t) less than St−bS_t - b, where v(t)v(t) is the selling rate in shares per day and η>0\eta > 0 is measured in currency unit days per share squared. The impact affects only the trades made at that moment and does not move later prices. Permanent impact is outside this model. The impact cost is η∫0hv(t)2 dt\eta \int_0^h v(t)^2\,dt, a deterministic amount. By the Cauchy-Schwarz inequality,

q2=(∫0hv(t) dt)2≤h∫0hv(t)2 dt,\begin{aligned} q^2 &= \Bigl(\int_0^h v(t)\,dt\Bigr)^2 \\[0.4em] &\le h \int_0^h v(t)^2\,dt, \end{aligned}

with equality only for a constant rate, so for a fixed horizon the constant rate v=q/hv = q/h has the smallest impact cost, ηq2/h\eta q^2/h. The expected cost of the constant-rate schedule is

E[C(h)]=bq+ηq2h.\mathbb{E}[C(h)] = bq + \frac{\eta q^2}{h} .

It decreases toward the spread floor bqbq as hh grows and becomes unbounded as h→0h \to 0, because linear impact makes very fast trading arbitrarily expensive.

In plain words Selling faster pushes the price further against each share, and at a constant rate the total impact cost is the push per share times the number of shares. Spreading the same sale over twice as long halves the speed and halves the impact cost.

Example With η=10−7\eta = 10^{-7} and q=100000q = 100000, the impact cost over one day is 10−7⋅1010/1=100010^{-7} \cdot 10^{10}/1 = 1000 currency units, so E[C(1)]=500+1000=1500\mathbb{E}[C(1)] = 500 + 1000 = 1500. Over two days the impact cost is 500500 and the expected cost 10001000. The units check, since currency unit days per share squared times shares squared per day leaves currency units.

Inventory variance

Definition In the model, the random part of the cost is −σ∫0hx(t) dWt-\sigma \int_0^h x(t)\,dW_t, an Itô integral of a deterministic integrand. By the Itô isometry of The Ito integral and isometry, its variance is

Var⁡(C)=σ2∫0hx(t)2 dt,\operatorname{Var}(C) = \sigma^2 \int_0^h x(t)^2\,dt,

and since the integrand is deterministic, CC is normally distributed. For the constant-rate schedule the inventory is x(t)=q(1−t/h)x(t) = q(1 - t/h), and the substitution u=t/hu = t/h gives

∫0hq2(1−th)2dt=q2h∫01(1−u)2 du=q2h3,\begin{aligned} &\int_0^h q^2\Bigl(1 - \frac{t}{h}\Bigr)^2 dt \\[0.4em] &\quad = q^2 h \int_0^1 (1 - u)^2\,du \\[0.4em] &\quad = \frac{q^2 h}{3}, \end{aligned}

so Var⁡(C(h))=σ2q2h/3\operatorname{Var}(C(h)) = \sigma^2 q^2 h/3, in currency units squared. It grows linearly with the horizon. Holding all qq shares for the time hh and selling them at the end would give the price variance σ2q2h\sigma^2 q^2 h, three times as much, because the constant-rate schedule reduces its exposure as it goes. The units of σ\sigma and hh have to match, here a square root of a day and days.

In plain words Every share not yet sold is exposed to price moves. The inventory variance adds up the squared exposure over the time it is carried, so a longer sale carries more price risk.

Example With σ=0.3\sigma = 0.3 and q=100000q = 100000, one day gives Var⁡(C)=0.09⋅1010/3=3×108\operatorname{Var}(C) = 0.09 \cdot 10^{10}/3 = 3 \times 10^{8} currency units squared, a standard deviation of about 17320.517320.5 currency units. Two days give 6×1086 \times 10^{8} and a standard deviation of about 24494.924494.9. A variance is in squared currency units, so it can be compared with a cost only after multiplying it by a coefficient in inverse currency units or taking its square root.

Liquidation horizon

Definition The liquidation horizon h>0h > 0 is the time over which the position is closed out. In the constant-rate model with a risk aversion λ≥0\lambda \ge 0, in inverse currency units, the horizon is chosen to minimize the mean-variance objective

J(h)=E[C]+λVar⁡(C)=bq+ηq2h+λσ2q2h3.\begin{aligned} J(h) &= \mathbb{E}[C] + \lambda \operatorname{Var}(C) \\[0.4em] &= bq + \frac{\eta q^2}{h} + \frac{\lambda \sigma^2 q^2 h}{3} . \end{aligned}

For η\eta, λ\lambda and σ\sigma all positive, J′(h)=−ηq2/h2+λσ2q2/3J'(h) = -\eta q^2/h^2 + \lambda \sigma^2 q^2/3 and J′′(h)=2ηq2/h3>0J''(h) = 2\eta q^2/h^3 > 0, so JJ is strictly convex on (0,∞)(0, \infty) with the unique minimizer

h∗=3ηλσ2,h^{*} = \sqrt{\frac{3\eta}{\lambda \sigma^2}} ,

which does not depend on qq in this model. Multiplying J′(h∗)=0J'(h^{*}) = 0 by h∗h^{*} shows that at the optimum the impact cost equals the risk penalty, so J(h∗)=bq+2q2ηλσ2/3J(h^{*}) = bq + 2q^2\sqrt{\eta \lambda \sigma^2/3}. The objective decreases before h∗h^{*} and increases after it, so if the feasible horizons form an interval, the best feasible horizon is h∗h^{*} moved to the nearer end of the interval when it lies outside. If λ=0\lambda = 0 or σ=0\sigma = 0, the objective decreases in hh and the formula for h∗h^{*} does not apply. The longest feasible horizon is then best, and without an upper bound no horizon is optimal, since the objective only approaches the spread floor bqbq. If η=0\eta = 0, the objective increases and the shortest feasible horizon is best. The optimum is over constant-rate schedules only. With λ\lambda and σ\sigma positive, the best of all deterministic schedules for a given horizon sells faster at the start, which lies outside this restricted model.

In plain words Waiting lowers the impact cost and raises the price risk, and the best horizon is where the two marginal effects cancel. More volatility makes waiting more expensive, so the horizon shortens.

Example With η=10−7\eta = 10^{-7}, λ=10−6\lambda = 10^{-6} and σ=0.3\sigma = 0.3, the optimal horizon is h∗=10/3≈1.82574h^{*} = \sqrt{10/3} \approx 1.82574 days. There the impact cost and the risk penalty are both about 547.72547.72, the expected cost is about 1047.721047.72, and J(h∗)≈1595.45J(h^{*}) \approx 1595.45, below J(1)=1800J(1) = 1800. If at most one day is allowed, the objective is still falling at one day, so the constrained choice is h=1h = 1. Doubling σ\sigma to 0.60.6 divides h∗h^{*} by 22, to about 0.912870.91287 days.

Liquidity adjustment

Definition A liquidity adjustment changes a mark-to-market value or a risk figure to account for the cost and the time of closing out. In this model the expected liquidation proceeds are

qS0−E[C(h)]=qS0−bq−ηq2h,\begin{aligned} &q S_0 - \mathbb{E}[C(h)] \\[0.4em] &\quad = q S_0 - bq - \frac{\eta q^2}{h}, \end{aligned}

so the adjustment to the mid mark is −E[C(h)]-\mathbb{E}[C(h)] for the chosen horizon. A risk-sensitive version uses the objective J(h)=E[C]+λVar⁡(C)J(h) = \mathbb{E}[C] + \lambda \operatorname{Var}(C), which is a decision criterion whose size depends on the chosen λ\lambda, not a quantile. Since CC is normal in this model, its quantile at a confidence level α\alpha is E[C]+Φ−1(α)Var⁡(C)\mathbb{E}[C] + \Phi^{-1}(\alpha)\sqrt{\operatorname{Var}(C)}, a different number. When a liquidity amount is added to a market risk figure, both have to be checked for the price risk of the inventory, which is counted twice if the market figure already covers price moves over the liquidation horizon and the liquidity amount includes the inventory variance. Jumps, changing depth or spreads, stochastic participation, price trends and permanent impact all change the distribution of the cost and lie outside this model.

In plain words The mark says what the position is worth at the mid price, and the adjustment says how much of that is expected to be lost in getting out. A risk-sensitive adjustment also charges for the uncertainty of the sale, at a rate the user chooses.

Example At the optimal horizon of about 1.8261.826 days the expected proceeds are qS0−1047.72qS_0 - 1047.72, so the mid mark is reduced by about 1047.721047.72 currency units. At one day the reduction is 15001500. The 9999 percent quantile of the one-day cost is

1500+2.3263⋅17320.5≈41794,\begin{aligned} &1500 + 2.3263 \cdot 17320.5 \\[0.4em] &\quad \approx 41794, \end{aligned}

more than twenty times J(1)=1800J(1) = 1800, which shows how different the two figures are.

Check yourself

  1. For the constructed position with q=100000q = 100000, b=0.005b = 0.005, η=10−7\eta = 10^{-7}, σ=0.3\sigma = 0.3 and λ=10−6\lambda = 10^{-6}, find the expected cost, the variance and the objective for a two-day constant-rate sale, and compare the objective with its minimum.
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The spread cost is 500500 and the impact cost is 10−7⋅1010/2=50010^{-7} \cdot 10^{10}/2 = 500, so E[C(2)]=1000\mathbb{E}[C(2)] = 1000. The variance is

Var⁡(C(2))=0.09⋅1010⋅23=6×108,\begin{aligned} \operatorname{Var}(C(2)) &= \frac{0.09 \cdot 10^{10} \cdot 2}{3} \\[0.4em] &= 6 \times 10^{8}, \end{aligned}

a standard deviation of about 24494.924494.9, and the risk penalty is 10−6⋅6×108=60010^{-6} \cdot 6 \times 10^{8} = 600. The objective is J(2)=1600J(2) = 1600, only about 4.554.55 above the minimum J(h∗)≈1595.45J(h^{*}) \approx 1595.45 at h∗≈1.826h^{*} \approx 1.826 days. The objective is flat near its minimum, while the expected cost at two days, 10001000, is lower than at the optimum because the optimum also pays for price risk.

  1. With the same inputs, the volatility doubles to σ=0.6\sigma = 0.6. Find the new optimal horizon and compare the objective there with the objective at the old horizon of about 1.8261.826 days.
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The volatility enters the optimal horizon squared in the denominator, so the horizon halves to

h∗=3⋅10−710−6⋅0.36≈0.91287\begin{aligned} h^{*} &= \sqrt{\frac{3 \cdot 10^{-7}}{10^{-6} \cdot 0.36}} \\[0.4em] &\approx 0.91287 \end{aligned}

days. There the impact cost and the risk penalty are both about 1095.451095.45, so the objective is about 500+2⋅1095.45=2690.89500 + 2 \cdot 1095.45 = 2690.89. Keeping the old horizon of about 1.8261.826 days under the new volatility gives the impact cost 547.72547.72 and the risk penalty 1200⋅1.826≈2190.891200 \cdot 1.826 \approx 2190.89, so J≈3238.61J \approx 3238.61. The faster sale is better because each day of exposure now costs four times as much in variance. A real stress could also widen the spread or raise η\eta, which is a different joint change and moves the balance in other ways.

  1. A colleague reports J(1)=1800J(1) = 1800 as the 9999 percent liquidation loss of the one-day sale. What is wrong with this, and what is the 9999 percent quantile of the cost in this model?
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The number 18001800 is the expected cost 15001500 plus λ\lambda times the variance, 10−6⋅3×108=30010^{-6} \cdot 3 \times 10^{8} = 300. It is a mean-variance criterion whose size depends on the chosen λ\lambda, and it has no confidence level. In this model the cost is normal with mean 15001500 and standard deviation about 17320.517320.5, so its 9999 percent quantile is about 1500+2.3263⋅17320.5≈417941500 + 2.3263 \cdot 17320.5 \approx 41794 currency units. The quantile depends on the horizon in another way too. Minimizing the 9999 percent quantile bq+ηq2/h+2.3263 σqh/3bq + \eta q^2/h + 2.3263\,\sigma q\sqrt{h/3} instead of JJ would select a horizon of only about 0.1350.135 days, with a quantile of about 2271222712. Both numbers hold only under the model’s assumptions of Brownian prices, constant spread and linear temporary impact.

How this lesson connects