Profit and loss attribution and sensitivities, financial engineering for quants lesson

A book of positions is marked at 142 currency units at the start of a short period and at 153.8 at its end, a gain of 11.8. How much of that came from the move in the underlying price, how much from the move in volatility and how much from the passage of time, and is anything left unexplained? Profit and loss attribution answers with a defined comparison and an expansion in sensitivities. This lesson derives the Taylor expansion of the change in value with its remainder, defines the market move, carry and the residual, and shows that the part of a joint move created by two factors together is assigned by a convention, not by the mathematics.

Key terms

Profit and loss attribution

Definition Let $V(t, x)$ be the signed mark of a fixed book of positions in one reporting currency, as a function of time $t$ in years and of a vector $x \in \mathbb{R}^d$ of risk factors. Over a period from $t$ to $t + \Delta t$ in which the factors move from $x$ to $x + \Delta x$ , the mark change is

\begin{aligned} \Delta V &= V(t + \Delta t, x + \Delta x) \\[0.4em] &\quad - V(t, x) . \end{aligned}

The total profit and loss of the period is the mark change plus the cash flows of the period, such as coupons received, financing paid and fees. A report that also covers new trades adds their effect, which the fixed book leaves out. Profit and loss attribution writes one chosen quantity of this kind, here the mark change, as named contributions plus a residual,

\Delta V = \sum_{k} A_k + R ,

where each $A_k$ is computed by a stated formula, for example a sensitivity times a factor move, and $R$ is defined as whatever remains. The decomposition is an identity by construction. Its content lies in the choice of the compared quantity, of the factors, of the formulas for the $A_k$ and of the convention that assigns joint effects. A formula for the mark change explains the mark change only, and cash flows enter only when they are added as their own terms.

In plain words Attribution first fixes exactly which change is being explained, for example the change in the mark of unchanged positions or the total gain including cash received. It then splits that change into pieces with names such as price, volatility and time, and the residual is the part the pieces do not cover.

Example Take the constructed value function, in currency units,

\begin{aligned} V(t, S, v) &= 0.01 S^2 + 2Sv \\[0.4em] &\quad + 50v^2 - 3t , \end{aligned}

where $S$ is a price in currency units, $v$ is a decimal volatility input, $t$ is in years and the coefficients carry the units that make every term a currency amount. At $t = 0$ , $S = 100$ and $v = 0.2$ the three market terms are $100$ , $40$ and $2$ , so the mark is $142$ . After $\Delta t = 0.01$ years with $\Delta S = 3$ and $\Delta v = 0.02$ the mark is $106.09 + 45.32 + 2.42 - 0.03 = 153.8$ , so $\Delta V = 11.8$ . The next cards split this into $11.6$ from first-order market terms, $0.23$ from second-order market terms and $-0.03$ from time, with residual $0$ . If the book had also received a coupon of $1.5$ in cash, its total profit and loss would be $13.3$ , and comparing that with the explained $11.8$ would produce a residual of $1.5$ that says nothing about pricing.

Market move

Definition The market move of the period is the vector $\Delta x$ of changes of the chosen risk factors over the period, each measured in a stated unit. The first-order contribution of factor $i$ is the sensitivity $\partial V/\partial x_i$ at $(t, x)$ times $\Delta x_i$ , and the market part of the change at fixed time is the full revaluation

\begin{aligned} \Delta_x V &= V(t, x + \Delta x) \\[0.4em] &\quad - V(t, x) . \end{aligned}

A sensitivity has the unit of value divided by the unit of its input, so the move must be expressed in the unit in which the sensitivity is a derivative. A derivative with respect to a decimal rate or a decimal volatility is multiplied by a decimal change, so one basis point is $\Delta x_i = 0.0001$ and one volatility point is $\Delta x_i = 0.01$ , while a sensitivity already quoted per basis point contains the factor $0.0001$ . The same care applies to currency translation of foreign sensitivities. The choice of factors is part of the attribution. The same move written in other coordinates, such as log prices in place of prices or variance in place of volatility, has different first-order terms, although the full revaluation is unchanged.

In plain words The market move is what the market did over the period, written in the coordinates the risk system uses. Its first-order effect is each sensitivity times the matching move, and the move has to be measured in the units of the sensitivity, or correct arithmetic gives a wrong attribution.

Example For the constructed function the sensitivities at the start are $\partial V/\partial S = 0.02S + 2v = 2.4$ and $\partial V/\partial v = 2S + 100v = 220$ . The moves $\Delta S = 3$ and $\Delta v = 0.02$ , which is two volatility points and not a relative rise of $2$ percent, give first-order contributions $2.4 \cdot 3 = 7.2$ and $220 \cdot 0.02 = 4.4$ , together $11.6$ . The full market revaluation at fixed time is $V(0, 103, 0.22) - V(0, 100, 0.2) = 153.83 - 142 = 11.83$ . Written with the variance $w = v^2$ as the volatility factor, the same move is $\Delta w = 0.0084$ and the sensitivity is $\partial V/\partial w = 550$ , so the first-order volatility term becomes $4.62$ instead of $4.4$ . The value is not a quadratic polynomial in $(S, w)$ , and the second-order expansion in these coordinates leaves a residual of about $0.0145$ against the same revalued $11.83$ .

Taylor expansion of PnL

Definition Fix the time $t$ and suppose $V(t, \cdot)$ is twice continuously differentiable on an open set that contains the segment from $x$ to $x + \Delta x$ . Write $\nabla_x V$ for its gradient and $H_x V$ for its Hessian in the market factors, and put

\begin{gathered} y_u = x + u\,\Delta x, \\[0.4em] g(u) = V(t, y_u), \quad 0 \le u \le 1 . \end{gathered}

The chain rule gives

\begin{aligned} g'(u) &= \nabla_x V(t, y_u)^{\top} \Delta x, \\[0.4em] g''(u) &= \Delta x^{\top} H_x V(t, y_u)\, \Delta x, \end{aligned}

and the one-variable Taylor formula with integral remainder, $g(1) = g(0) + g'(0) + \int_0^1 (1 - u)\, g''(u)\,du$ , gives

\begin{aligned} \Delta_x V &= \nabla_x V^{\top} \Delta x \\[0.4em] &\quad + \tfrac12 \Delta x^{\top} H_x V\, \Delta x + R_3 , \end{aligned}

with the derivatives taken at $(t, x)$ and

\begin{aligned} R_3 &= \int_0^1 (1 - u) \\[0.4em] &\qquad \times \bigl(g''(u) - g''(0)\bigr)\,du . \end{aligned}

Continuity of the Hessian gives $R_3/\lVert \Delta x \rVert^2 \to 0$ as $\Delta x \to 0$ , a local statement that gives no number for a move of a given size. If in addition $V(t, \cdot)$ is three times continuously differentiable there and $\lvert g'''(u) \rvert \le M \lVert \Delta x \rVert^3$ for $0 \le u \le 1$ , then $R_3 = \int_0^1 \tfrac12 (1 - u)^2 g'''(u)\,du$ and

\lvert R_3 \rvert \le \frac{M}{6}\,\lVert \Delta x \rVert^3 .

The delta-gamma-theta expansion adds a first-order time term,

\begin{aligned} \Delta V &\approx V_t\,\Delta t + \nabla_x V^{\top} \Delta x \\[0.4em] &\quad + \tfrac12 \Delta x^{\top} H_x V\, \Delta x , \end{aligned}

where $V_t = \partial V/\partial t$ at $(t, x)$ is theta per year. It keeps market curvature but omits the time curvature $\tfrac12 V_{tt}\,\Delta t^2$ and the time-market cross terms $\Delta t\, \nabla_x V_t^{\top} \Delta x$ , which a full second-order expansion in the joint coordinates $(t, x)$ contains. It is exact when $V(t, x) = a + bt + Q(x)$ with constants $a$ and $b$ and a polynomial $Q$ of degree at most two. A kink of $V$ on the segment, as at the strike of an expiring option, breaks the hypothesis, and the expansion then has no basis.

In plain words Along the straight line from the old factor values to the new ones the value is a function of one variable. Its slope at the start gives the sensitivity terms, its curvature at the start gives the second-order terms, and the change of curvature along the way is the remainder. How large that remainder is for an actual move depends on how fast the curvature changes, which the Hessian at the start does not reveal.

Example For the constructed function the Hessian in $(S, v)$ is constant,

H_x V = \begin{pmatrix} 0.02 & 2 \\ 2 & 100 \end{pmatrix},

so the second-order market term is

\begin{aligned} &\tfrac12 \cdot 0.02 \cdot 3^2 + 2 \cdot 3 \cdot 0.02 \\[0.4em] &\quad + \tfrac12 \cdot 100 \cdot 0.02^2 \\[0.4em] &= 0.09 + 0.12 + 0.02 = 0.23 . \end{aligned}

The cross term $0.12$ has no factor one half because the symmetric Hessian contains the mixed derivative twice. With $V_t = -3$ the time term is $-3 \cdot 0.01 = -0.03$ , and the expansion gives $11.6 + 0.23 - 0.03 = 11.8$ , exactly the revalued change, because the function has the form $a + bt + Q(S, v)$ with $Q$ quadratic.

Carry

Definition Carry is the profit and loss from the passage of time over the period under a stated convention for what an unchanged market means, together with whatever accruals and financing flows of the period the comparison includes. With the factors held fixed in the stated coordinates, the mark part of carry is

\begin{aligned} C &= V(t + \Delta t, x) \\[0.4em] &\quad - V(t, x), \end{aligned}

and its first-order estimate is $V_t\,\Delta t$ , theta per year times the elapsed time in years. The number depends on the convention. For a bond, holding its own yield fixed, holding the zero curve fixed as a function of time to maturity and letting today’s forward rates be realized are three different unchanged markets with three different carries, and the market move is then measured from the chosen baseline, so the total change is the same. A carry computed from theta also depends on the day count that turns calendar days into $\Delta t$ .

In plain words Carry is what the book would have earned if nothing in the market had moved. Since an unchanged market can be read in several ways, the reading has to be fixed before the number means anything.

Example For the constructed function the carry is exactly $-3 \cdot 0.01 = -0.03$ , since time enters linearly and without interaction. For a second constructed case, take a zero-coupon bond that pays $100$ in $5$ years, with continuously compounded zero rates $y(\tau) = 0.03 + 0.002\tau$ for time to maturity $\tau$ in years, so its value is $100e^{-0.04 \cdot 5} \approx 81.8731$ . Over half a year with its own yield fixed at $0.04$ the value becomes $100e^{-0.18} \approx 83.5270$ , a carry of about $1.6539$ . If the curve instead keeps its shape in time to maturity, the bond is discounted at $y(4.5) = 0.039$ and is worth about $83.9037$ , a carry of about $2.0307$ , which includes about $0.3767$ of roll-down. If today’s forward rates are realized, the bond grows at the half-year zero rate $y(0.5) = 0.031$ to about $83.1520$ , a carry of about $1.2789$ . Theta with the own yield fixed is $0.04 \cdot 81.8731 \approx 3.2749$ per year, and $3.2749 \cdot 0.5 \approx 1.6375$ is the first-order estimate of the first carry.

Attribution residual

Definition The attribution residual is the compared change minus the sum of the explanatory terms,

R = \Delta V - \sum_{k} A_k .

It is defined by subtraction, so it collects everything the chosen terms leave out. Its sources include terms of higher order than the expansion, a kink or fast-changing curvature on the path of the move, cash flows or position changes that the compared quantity contains but the terms do not, a change of model, calibration or market data between the two marks, and factors that moved but are not in $x$ . A residual is therefore not automatically a gain unexplained by risk. A small residual shows that the chosen terms explained this comparison, not that every risk of the book is modeled.

In plain words The residual is the gap between what happened and what the formula explains. Its size and sign are a diagnostic, and finding its source means asking which hypothesis behind the explained terms failed.

Example Add a cubic term $cS^3$ with $c = 0.0001$ to the constructed function, which raises the initial mark to $242$ . The sensitivities at the start become $\partial V/\partial S = 2.4 + 3cS^2 = 5.4$ and $\partial^2 V/\partial S^2 = 0.02 + 6cS = 0.08$ , and the expansion now explains

\begin{aligned} &5.4 \cdot 3 + 4.4 + 0.36 \\[0.4em] &\quad + 0.12 + 0.02 - 0.03 = 21.07 . \end{aligned}

Revaluation gives a change of $263.0727 - 242 = 21.0727$ , so the residual is $0.0027$ . It is exactly $c\,\Delta S^3 = 0.0001 \cdot 27$ , the one term of $c(S + \Delta S)^3$ beyond second order. The remainder formula gives the same number, since $g'''(u) = 6c\,\Delta S^3 = 0.0162$ is constant and $\int_0^1 \tfrac12 (1 - u)^2\,du = \tfrac16$ . The residual reveals a missing order of the expansion, not an error in the quadratic identity.

Attribution convention

Definition An attribution convention is the rule that assigns each part of the change to a named column. A sequential convention moves the factors one at a time in a fixed order and revalues after each step. If $\Delta x^{(i)}$ moves the first $i$ factors of the order and leaves the others at their old values, with $\Delta x^{(0)} = 0$ , the $i$ th factor of the order receives

\begin{aligned} &V(t, x + \Delta x^{(i)}) \\[0.4em] &\quad - V(t, x + \Delta x^{(i-1)}) . \end{aligned}

The steps telescope to $\Delta_x V$ for every order, but the individual columns depend on the order whenever the value has interactions, that is nonzero mixed derivatives between the factors. Other conventions average the sequential contributions over all orders, the Shapley allocation, which for two factors splits the interaction equally, or report the interaction in its own column. The value function determines the size of the interaction and the convention determines only its label, so contributions from different reports can be compared only under the same convention.

In plain words When two factors move together, part of the change exists only because both moved, and no single factor owns it. The convention decides where that part is shown, while the total stays the same.

Example In the constructed example the interaction is the cross term $2\,\Delta S\,\Delta v = 0.12$ . Moving $S$ first gives $V(0, 103, 0.2) - 142 = 149.29 - 142 = 7.29$ to price and then $153.83 - 149.29 = 4.54$ to volatility, because the volatility move is evaluated at the higher price. Moving $v$ first gives $146.42 - 142 = 4.42$ to volatility and $153.83 - 146.42 = 7.41$ to price. Both orders add to $11.83$ , and the columns differ by exactly $0.12$ . The equal split reports $7.35$ and $4.48$ , and a separate interaction column reports $7.29$ , $4.42$ and $0.12$ . Time enters this function without interactions, so the carry of $-0.03$ is the same in every order.

Check yourself

For the constructed function at $t = 0$ , $S = 100$ and $v = 0.2$ , the price falls by $2$ and volatility rises by one point, $\Delta v = 0.01$ , over $0.01$ years. Compute the first-order, second-order and carry terms, compare them with revaluation, and find where the interaction goes under each sequential order.

Show the answer

The first-order terms are $2.4 \cdot (-2) = -4.8$ and $220 \cdot 0.01 = 2.2$ , together $-2.6$ . The second-order terms are

\begin{aligned} &\tfrac12 \cdot 0.02 \cdot 4 + 2 \cdot (-2) \cdot 0.01 \\[0.4em] &\quad + \tfrac12 \cdot 100 \cdot 0.01^2 \\[0.4em] &= 0.04 - 0.04 + 0.005 = 0.005, \end{aligned}

and the carry is $-3 \cdot 0.01 = -0.03$ , so the expansion gives $-2.625$ . Revaluation gives $V(0.01, 98, 0.21) = 96.04 + 41.16 + 2.205 - 0.03 = 139.375$ , a change of $-2.625$ , exact again because the function is quadratic in the market factors and linear in time. The interaction is $-0.04$ , negative because the factors moved in opposite directions while the mixed derivative is positive. Moving the price first gives $-4.76$ to price and $2.165$ to volatility, so volatility receives the interaction. Moving volatility first gives $2.205$ to volatility and $-4.8$ to price, so price receives it.

For the same function a report says that volatility rose by $2$ percent, and a colleague multiplies the sensitivity $220$ by $2$ . Give the first-order volatility term under the two readings of the report and explain the error. For a constructed zero-coupon bond worth $100e^{-5y}$ at the continuously compounded yield $y = 0.04$ , give the first-order effect of a rise of $25$ basis points.

Show the answer

The sensitivity $220$ is per unit of decimal volatility, so the move must be a decimal change. Read as two volatility points, the move is $\Delta v = 0.02$ and the term is $220 \cdot 0.02 = 4.4$ , against $4.42$ from revaluation in $v$ alone. Read as a relative rise of $2$ percent, the move is $\Delta v = 0.2 \cdot 0.02 = 0.004$ and the term is $0.88$ , against $0.8808$ from revaluation. The product $220 \cdot 2 = 440$ treats $2$ as a decimal change, a move of $200$ volatility points. The report has to say which reading it means. For the bond, $\partial V/\partial y = -5 \cdot 100e^{-0.2} \approx -409.37$ per unit of decimal yield, which is about $-0.0409$ per basis point. A rise of $25$ basis points is $\Delta y = 0.0025$ , so the first-order effect is

\begin{aligned} &-409.37 \cdot 0.0025 \\[0.4em] &\quad \approx -1.0234, \end{aligned}

close to the revalued $-1.0170$ , while multiplying by $25$ would give about $-10234$ .

A constructed position has mark $10(S - 100)^{+}$ over the period, with no time dependence, and the price is $99$ . The risk system reports a delta of $0$ and a gamma of $0$ , and the price rises by $3$ . What does the Taylor attribution explain, what is the residual, and which hypothesis fails? What changes for a rise of $0.5$ ?

Show the answer

At $S = 99$ the mark is $0$ and its first two derivatives are $0$ , so the expansion explains $0$ . Revaluation at $102$ gives $10 \cdot 2 = 20$ , so the residual is $20$ , the whole change. The expansion needs the value to be twice continuously differentiable on an open set containing the segment from $99$ to $102$ , and the mark has a kink at $100$ , where it is not differentiable. For a rise of $0.5$ the segment from $99$ to $99.5$ stays where the mark is identically $0$ , so the expansion is exact and the residual is $0$ . Full revaluation handles the larger move, or the path can be split at the strike, with $0$ from $99$ to $100$ and $10 \cdot 2 = 20$ from $100$ to $102$ using the right derivative $10$ . An option shortly before expiry has a smooth value, but its gamma changes fast near the strike, so its residual for a move across the strike can be large for a related reason, since the curvature at the start says little about the curvature along the path.

Profit and loss attribution and sensitivities

Key terms

Profit and loss attribution

Market move

Taylor expansion of PnL

Carry

Attribution residual

Attribution convention

Check yourself

How this lesson connects

Builds on

Used later in