Changing Measure to Compute Drifted Brownian Tail Probabilities

Why Change Measure for a Brownian Tail?

For a Brownian motion with constant drift, the finite-time distribution is Gaussian, so its tail probability can be computed directly. A change of measure gives a second route: it absorbs the drift into the probability law, leaving a standard Brownian terminal value and a density weight. This is useful not because it avoids the Gaussian calculation, but because it makes the roles of the drift, the event, and the likelihood ratio explicit. The same bookkeeping appears in more complicated measure-change arguments.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$, and fix a finite horizon $T>0$. Let $\mu\in\mathbb{R}$ be constant and define the drifted process $X_t=B_t+\mu t$. We compute $\mathbb{P}(X_T>a)$ for a real threshold $a$. The strict or non-strict inequality gives the same probability here, because $X_T$ has a continuous Gaussian distribution.

Definition: The drifted Brownian upper-tail probability at time $T$ and threshold $a$ is $\mathbb{P}(B_T+\mu T>a)$. The parameter $\mu$ is the drift coefficient; the diffusion coefficient of $X$ is $1$.

Constructing the New Measure

Define, for $0\le t\le T$, the process

Z_t=\exp\left(-\mu B_t-\frac12\mu^2t\right).

It is the stochastic exponential of the Ito integral $-\int_0^t\mu\,dB_s$. For every finite horizon $T$, the integrand $H_s=-\mu$ is adapted and progressively measurable, and $\mathbb{E}\int_0^T H_s^2\,ds=\mu^2T<\infty$; thus it belongs to the class $\mathcal{L}^2$ of progressively measurable integrands with finite expected integrated square. Ito's formula applied to the exponential gives $dZ_t=-\mu Z_t\,dB_t$. This integral representation alone gives a local martingale; to use $Z_T$ as a probability density, we must verify that it is a true martingale with expectation one.

Theorem: For constant $\mu$ and finite $T$, $Z_t$ is a positive martingale on $[0,T]$ with $\mathbb{E}[Z_t]=1$. Define $\mathbb{Q}$ on $\mathcal{F}_T$ by $d\mathbb{Q}/d\mathbb{P}=Z_T$. Then under $\mathbb{Q}$, $B_t^{\mathbb{Q}}=B_t+\mu t$ is a standard Brownian motion on $[0,T]$.

Proof. For $0\le s\le t\le T$, write $B_t=B_s+(B_t-B_s)$. Brownian increments are independent of $\mathcal{F}_s$ and $B_t-B_s$ has distribution $N(0,t-s)$. Therefore

Z_t=Z_s\exp\left(-\mu(B_t-B_s)-\frac12\mu^2(t-s)\right)

and the conditional expectation of the second factor given $\mathcal{F}_s$ is $1$. Indeed, for $G\sim N(0,t-s)$, its normal moment-generating function gives $\mathbb{E}[\exp(-\mu G)]=\exp(\tfrac12\mu^2(t-s))$, which cancels the deterministic factor. Hence $\mathbb{E}[Z_t\mid\mathcal{F}_s]=Z_s$. Also $\mathbb{E}[Z_0]=1$, so $Z$ is a martingale and $\mathbb{E}[Z_t]=1$ for every $t\le T$. It is strictly positive, so $Z_T$ defines a probability measure $\mathbb{Q}$ equivalent to $\mathbb{P}$ on $\mathcal{F}_T$. Novikov's condition is satisfied as well: $\mathbb{E}[\exp(\tfrac12\int_0^T\mu^2\,ds)]=\exp(\tfrac12\mu^2T)<\infty$. By Girsanov's theorem, under $\mathbb{Q}$ the process $B_t+\mu t$ is standard Brownian motion. $\square$

Girsanov's theorem is the standard measure-change result used here; its proof is not repeated. In this application, its hypotheses have been checked: the kernel is the constant $-\mu$, its integrated square is finite, and Novikov's condition holds. Notice the direction of the change: $Z_T$ is $d\mathbb{Q}/d\mathbb{P}$, whereas the desired probability is under $\mathbb{P}$. We will therefore need the reciprocal density to express that probability as an expectation under $\mathbb{Q}$.

The Tail Formula from the Density

By the definition of $\mathbb{Q}$, the density process through time $t$ is $Z_t=d\mathbb{Q}/d\mathbb{P}|_{\mathcal{F}_t}$. Since $Z_t>0$, the reciprocal is $d\mathbb{P}/d\mathbb{Q}|_{\mathcal{F}_t}=1/Z_t$. Under $\mathbb{Q}$, $B_T^{\mathbb{Q}}=B_T+\mu T=X_T$, so the drifted event becomes a standard Brownian event. Rewriting the reciprocal density in terms of $B_T^{\mathbb{Q}}$ gives

\frac{d\mathbb{P}}{d\mathbb{Q}}\Big|_{\mathcal{F}_T} =\exp\left(\mu B_T+\frac12\mu^2T\right) =\exp\left(\mu B_T^{\mathbb{Q}}-\frac12\mu^2T\right).

Result: For $T>0$, $\mu,a\in\mathbb{R}$, and $X_T=B_T+\mu T$, $$ \mathbb{P}(X_T>a)=\Phi\left(\frac{\mu T-a}{\sqrt{T}}\right), $$ where $\Phi$ is the standard normal cumulative distribution function.

Proof. Let $Y=B_T^{\mathbb{Q}}$, which under $\mathbb{Q}$ has distribution $N(0,T)$. The event $X_T>a$ is exactly $Y>a$. Changing from $\mathbb{P}$ to $\mathbb{Q}$ with the reciprocal density therefore gives

\mathbb{P}(X_T>a) =\mathbb{E}^{\mathbb{Q}}\left[ \exp\left(\mu Y-\frac12\mu^2T\right)\mathbf{1}_{\{Y>a\}} \right].

The expectation is finite: without the indicator, the expectation of the positive exponential under $N(0,T)$ is $1$. Writing out the normal density and completing the square,

\exp\left(\mu y-\frac12\mu^2T\right) \frac{1}{\sqrt{2\pi T}}\exp\left(-\frac{y^2}{2T}\right) = \frac{1}{\sqrt{2\pi T}}\exp\left(-\frac{(y-\mu T)^2}{2T}\right).

Thus the expectation is the integral of the $N(\mu T,T)$ density over $y>a$. With $u=(y-\mu T)/\sqrt{T}$, the lower integration limit is $(a-\mu T)/\sqrt{T}$, and the upper tail is $1-\Phi((a-\mu T)/\sqrt{T})=\Phi((\mu T-a)/\sqrt{T})$. This proves the formula. $\square$

The equality with the Gaussian tail is an independent check: under $\mathbb{P}$, $B_T\sim N(0,T)$, and therefore $X_T\sim N(\mu T,T)$. Its upper tail at $a$ has precisely the expression just derived. This check also confirms that the sign of the drift in the final numerator must be positive.

Worked Calculations

Worked Example: Positive Drift and a Zero Threshold

Take $\mu=1$, $T=1$, and $a=0$. Compute $\mathbb{P}(B_1+1>0)$ by the change of measure.

Under $\mathbb{Q}$, $Y=B_1^{\mathbb{Q}}=B_1+1$ has distribution $N(0,1)$. The reciprocal density at time $1$ is $\exp(Y-\tfrac12)$, so

\mathbb{P}(B_1+1>0) =\int_0^\infty e^{y-1/2}\frac{1}{\sqrt{2\pi}}e^{-y^2/2}\,dy =\int_0^\infty\frac{1}{\sqrt{2\pi}}e^{-(y-1)^2/2}\,dy =\Phi(1).

For the independent check, under $\mathbb{P}$, $B_1+1\sim N(1,1)$. Its standardized threshold is $(0-1)/1=-1$, so the upper tail is $1-\Phi(-1)=\Phi(1)$, as obtained by weighting under $\mathbb{Q}$.

Worked Example: Negative Drift and a Positive Threshold

Take $\mu=-\tfrac12$, $T=4$, and $a=1$. Here $\mu T=-2$ and $\sqrt{T}=2$. Under $\mathbb{Q}$, $Y=B_4^{\mathbb{Q}}=B_4-2$ has distribution $N(0,4)$, and the reciprocal density is $\exp(-\tfrac12Y-\tfrac12)$, since $\tfrac12\mu^2T=\tfrac12\cdot\tfrac14\cdot4=\tfrac12$. Thus

\begin{aligned} \mathbb{P}(B_4-2>1) &=\int_1^\infty e^{-y/2-1/2}\frac{1}{\sqrt{8\pi}}e^{-y^2/8}\,dy\\ &=\int_1^\infty\frac{1}{\sqrt{8\pi}}e^{-(y+2)^2/8}\,dy\\ &=1-\Phi\left(\frac{1+2}{2}\right) =1-\Phi\left(\frac32\right). \end{aligned}

For the check under $\mathbb{P}$, the drifted terminal value has mean $-2$ and standard deviation $2$. Standardizing the threshold $1$ gives $(1-(-2))/2=3/2$, so its upper tail is also $1-\Phi(3/2)$. A negative drift makes this positive-threshold event less likely than a zero-drift event, consistent with the result.

Worked Example: A Positive Drift with a Nonzero Threshold

Take $\mu=\tfrac34$, $T=4$, and $a=1$. Then $\mu T=3$, $\sqrt{T}=2$, and $\tfrac12\mu^2T=\tfrac12\cdot\tfrac{9}{16}\cdot4=\tfrac98$. Under $\mathbb{Q}$, $Y=B_4^{\mathbb{Q}}\sim N(0,4)$, and the reciprocal density is $\exp(\tfrac34Y-\tfrac98)$. Completing the square gives

\begin{aligned} \mathbb{P}(B_4+3>1) &=\int_1^\infty e^{3y/4-9/8}\frac{1}{\sqrt{8\pi}}e^{-y^2/8}\,dy\\ &=\int_1^\infty\frac{1}{\sqrt{8\pi}}e^{-(y-3)^2/8}\,dy\\ &=1-\Phi\left(\frac{1-3}{2}\right) =1-\Phi(-1)=\Phi(1). \end{aligned}

Directly under $\mathbb{P}$, $B_4+3$ has mean $3$ and standard deviation $2$. The standardized threshold is $(1-3)/2=-1$, giving $1-\Phi(-1)=\Phi(1)$. Both calculations agree; the positive drift moves the mean above the threshold.

Common Mistakes and Checks

Reversing the density. Since $Z_T=d\mathbb{Q}/d\mathbb{P}$, a $\mathbb{P}$-probability written as a $\mathbb{Q}$-expectation uses $1/Z_T=d\mathbb{P}/d\mathbb{Q}$, not $Z_T$. For example, with $\mu=1,T=1,a=0$, using $Z_T$ as the weight would give $\mathbb{E}^{\mathbb{Q}}[e^{-Y-1/2}\mathbf{1}_{\{Y>0\}}]$, which equals $\int_0^\infty\phi(y+1)\,dy=1-\Phi(1)$, the wrong tail. The correct reciprocal weight is $e^{Y-1/2}$, giving $\Phi(1)$.

Changing the event without changing the threshold. Under $\mathbb{Q}$, the event is $B_T^{\mathbb{Q}}>a$, because $B_T^{\mathbb{Q}}=B_T+\mu T=X_T$. It is not $B_T^{\mathbb{Q}}>a-\mu T$. The shift appears instead in the density and, after completing the square, in the shifted normal density.

Confusing a local martingale with a true martingale. The stochastic differential $dZ_t=-\mu Z_t\,dB_t$ by itself gives a local martingale conclusion. In this constant-kernel, finite-time setting, the conditional Gaussian calculation verifies the true martingale property and expectation one directly; Novikov's condition is also finite. That verification is what allows $Z_T$ to define a probability measure. The resulting tail formula is under $\mathbb{P}$; $\mathbb{Q}$ is only the computational measure.

As a final limiting check, set $\mu=0$. Then $Z_t=1$, the two measures coincide, and the formula reduces to $\mathbb{P}(B_T>a)=\Phi(-a/\sqrt{T})$. This is the standard Brownian upper tail and confirms that no drift correction remains when the drift vanishes.

Check Your Understanding

For each question, assume $T>0$, and use the density in the direction required by the expectation.

For general constant $\mu$, write $Z_t=d\mathbb{Q}/d\mathbb{P}|_{\mathcal{F}_t}$ and express $d\mathbb{P}/d\mathbb{Q}|_{\mathcal{F}_t}$ in terms of $B_t^{\mathbb{Q}}$.
Compute $\mathbb{P}(B_2+\tfrac12\cdot2>2)$ by substituting the parameters into the drifted-tail formula.
Explain why the event $B_T+\mu T>a$ becomes $B_T^{\mathbb{Q}}>a$ under the chosen measure.
Complete the square in $e^{\mu y-\mu^2T/2}\exp(-y^2/(2T))$ and identify the resulting normal mean and variance.
For $\mu=0$, verify that the change of measure is trivial and recover the standard Brownian upper-tail probability.