Friday 4 September 2026

Setting up the measure change

Under $\mathbb{P}$, $W$ is a standard Brownian motion and $X_t=\nu t+W_t$ carries drift $\nu$. We want a measure under which the drift is removed. Define, for each $t$, the positive random variable $$L_t=\exp (-\nu W_t-\frac{1}{2}{\nu }^{2}t),$$ which is an $({\mathcal{F}}_t,\mathbb{P})$–martingale with ${\mathbb{E}}^{\mathbb{P}}[L_t]=1$ (Novikov's condition holds trivially, the exponent being linear in $W$). Set $${\frac{d\mathbb{Q}}{d\mathbb{P}}|}_{{\mathcal{F}}_t}=L_t.$$

By Girsanov's theorem, under $\mathbb{Q}$ the process $${\tilde{W}}_t:=W_t+\nu t=X_t$$ is a standard Brownian motion. So under $\mathbb{Q}$, $X$ is driftless standard Brownian motion — exactly the process whose passage law we were handed.

The derivative we actually need runs the other way. Since $L_t$ is invertible, $${\frac{d\mathbb{P}}{d\mathbb{Q}}|}_{{\mathcal{F}}_t}=L_t^{-1}=\exp (\nu W_t+\frac{1}{2}{\nu }^{2}t).$$ Substituting $W_t=X_t-\nu t$ gives the required form in terms of $X$: $${\frac{d\mathbb{P}}{d\mathbb{Q}}|}_{{\mathcal{F}}_t}=Z_t,Z_t:=\exp (\nu X_t-\frac{1}{2}{\nu }^{2}t).$$ Note $Z_t$ is a $\mathbb{Q}$–martingale (it is the Girsanov density of $\mathbb{P}$ relative to $\mathbb{Q}$, and under $\mathbb{Q}$ the process $X$ is a $\mathbb{Q}$–BM, so $Z$ is a stochastic exponential of $\nu X$).

The passage-time density

Let $f\ge 0$ be Borel. For a bounded horizon $T$, the stopping time $\tau \wedge T$ is bounded, and the density process satisfies the optional-sampling identity for Radon–Nikodym derivatives: $${\frac{d\mathbb{P}}{d\mathbb{Q}}|}_{{\mathcal{F}}_{\tau \wedge T}}=Z_{\tau \wedge T}.$$ Hence, since $\{\tau \le T\}\in {\mathcal{F}}_{\tau \wedge T}$, $${\mathbb{E}}^{\mathbb{P}}[f(\tau ){\mathbf{1}}_{\{\tau \le T\}}]={\mathbb{E}}^{\mathbb{Q}}[Z_{\tau \wedge T}f(\tau ){\mathbf{1}}_{\{\tau \le T\}}]={\mathbb{E}}^{\mathbb{Q}}[Z_{\tau }f(\tau ){\mathbf{1}}_{\{\tau \le T\}}].$$

This is the step a candidate is most likely to fumble: you must weight by $Z$ evaluated at $\tau$, not at $T$. Using a bounded stopping time $\tau \wedge T$ (rather than the fixed $T$) is precisely what lets you do this cleanly, sidestepping any uniform-integrability worry about $Z_T$. On the event $\{\tau <\infty \}$ the definition of $\tau$ pins the path: $X_{\tau }=a$, so $$Z_{\tau }=\exp (\nu a-\frac{1}{2}{\nu }^2\tau )\text{on }\{\tau <\infty \}.$$ Letting $T\uparrow \infty$ (monotone convergence, $f\ge 0$), $${\mathbb{E}}^{\mathbb{P}}[f(\tau ){\mathbf{1}}_{\{\tau <\infty \}}]={\mathbb{E}}^{\mathbb{Q}}[e^{\nu a-\frac{1}{2}{\nu }^2\tau }f(\tau ){\mathbf{1}}_{\{\tau <\infty \}}].$$

Under $\mathbb{Q}$, $\tau$ is the first passage of a standard Brownian motion to $a$, so it has the given driftless density and is finite $\mathbb{Q}$–a.s. Therefore $$\mathbb{P}(\tau \in dt)=e^{\nu a-\frac{1}{2}{\nu }^{2}t}·\frac{a}{\sqrt{2\pi t^3}}{e}^{-a^2/(2t)}dt.$$ Complete the square in the exponent: $$\nu a-\frac{1}{2}{\nu }^{2}t-\frac{a^2}{2t}=-\frac{1}{2t}(a^2-2\nu at+{\nu }^2{t}^2)=-\frac{(a-\nu t{)}^2}{2t}.$$ Hence $$\boxed{\mathbb{P}(\tau \in dt)=\frac{a}{\sqrt{2\pi t^3}}\exp (-\frac{(a-\nu t{)}^2}{2t})dt}$$ the inverse-Gaussian first-passage law.

Laplace transform and total mass

Take $f(t)=e^{-\lambda t}$, $\lambda >0$. Then $${\mathbb{E}}^{\mathbb{P}}\left[e^{-\lambda \tau }{\mathbf{1}}_{\{\tau <\infty \}}\right]=e^{\nu a}{\mathbb{E}}^{\mathbb{Q}}\left[e^{-(\lambda +\frac{1}{2}{\nu }^2)\tau }\right]=e^{\nu a}{e}^{-a\sqrt{2(\lambda +{\nu }^2/2)}},$$ using the driftless Laplace transform ${\mathbb{E}}^{\mathbb{Q}}[e^{-s\tau }]=e^{-a\sqrt{2s}}$ with $s=\lambda +\frac{1}{2}{\nu }^2$. Thus $$\boxed{{\mathbb{E}}^{\mathbb{P}}\left[e^{-\lambda \tau }{\mathbf{1}}_{\{\tau <\infty \}}\right]=\exp \left(a\right(\nu -\sqrt{{\nu }^2+2\lambda }\left)\right)}.$$

Letting $\lambda \downarrow 0$ and using monotone convergence recovers the total mass: $$\mathbb{P}(\tau <\infty )=\exp (a(\nu -|\nu |))=\{\begin{array}{cc}1,\hfill & \nu \ge 0,\hfill \\ [\mathrm{2pt}]e^{2\nu a},\hfill & \nu <0.\hfill \end{array}$$

This is the sanity check: with upward or zero drift the level $a>0$ is hit with certainty, whereas with downward drift ($\nu <0$) the hitting probability is $e^{2\nu a}<1$, decaying in the barrier $a$ and in the strength of the adverse drift. The Laplace transform at $\lambda =0$ agrees with this directly.

Note

The slicker but less illuminating route to part 3 skips the density: the process $M_t=\exp (\theta X_t-(\theta \nu +\frac{1}{2}{\theta }^2)t)$ is a $\mathbb{P}$–martingale for any $\theta$, and optional stopping at $\tau$ (with $\theta >0$ chosen so that $\theta \nu +\frac{1}{2}{\theta }^2=\lambda$, i.e. $\theta =\sqrt{{\nu }^2+2\lambda }-\nu$) gives $\mathbb{E}[e^{-\lambda \tau }{\mathbf{1}}_{\{\tau <\infty \}}]e^{\theta a}=1$, reproducing the boxed answer. The Girsanov derivation is worth doing because it delivers the full density in one stroke. A common wrong turn is to weight by $Z_T=e^{\nu X_T-\frac{1}{2}{\nu }^{2}T}$ with the fixed horizon and to forget that on the passage event the relevant value is $X_{\tau }=a$; that error typically produces a stray dependence on $X_T$ that cannot be integrated out.

Throughout, a population regression coefficient of $A$ on $B$ (mean zero) is $\mathrm{C}\mathrm{o}\mathrm{v}(A,B)/\mathrm{V}\mathrm{a}\mathrm{r}(B)$; this is the $L^2$ projection, exact here because everything is jointly Gaussian. For Gaussian vectors a partial covariance $\mathrm{C}\mathrm{o}\mathrm{v}(A,B\mid C)$ equals the covariance of the residuals of $A$ and $B$ after linearly regressing each on $C$ — this is the content of the Frisch–Waugh–Lovell identity, and it is the step that makes part 3 mechanical.

First record the second moments we will reuse:

$$\mathrm{V}\mathrm{a}\mathrm{r}(Z)=\mathrm{V}\mathrm{a}\mathrm{r}(C)+\mathrm{V}\mathrm{a}\mathrm{r}({ϵ}_Z)=1+1=2,$$

$$\mathrm{V}\mathrm{a}\mathrm{r}(X)=\mathrm{V}\mathrm{a}\mathrm{r}(Z)+\mathrm{V}\mathrm{a}\mathrm{r}(U)+\mathrm{V}\mathrm{a}\mathrm{r}({ϵ}_X)=2+1+1=4,$$

$$\mathrm{C}\mathrm{o}\mathrm{v}(X,U)=\mathrm{V}\mathrm{a}\mathrm{r}(U)=1,\mathrm{C}\mathrm{o}\mathrm{v}(X,C)=\mathrm{C}\mathrm{o}\mathrm{v}(Z,C)=1,\mathrm{C}\mathrm{o}\mathrm{v}(Z,C)=1.$$

Part 1: naive OLS

$$\mathrm{C}\mathrm{o}\mathrm{v}(X,Y)=2\mathrm{V}\mathrm{a}\mathrm{r}(X)+2\mathrm{C}\mathrm{o}\mathrm{v}(X,U)+3\mathrm{C}\mathrm{o}\mathrm{v}(X,C)=2(4)+2(1)+3(1)=13.$$

$${\beta }_{\mathrm{O}\mathrm{L}\mathrm{S}}=\frac{\mathrm{C}\mathrm{o}\mathrm{v}(X,Y)}{\mathrm{V}\mathrm{a}\mathrm{r}(X)}=\frac{13}{4}=3.25.$$

The bias is $3.25-2=1.25$, contributed by the two open backdoor paths $X\leftarrow U\to Y$ and $X\leftarrow Z\leftarrow C\to Y$; numerically $(2·1+3·1)/4=5/4$.

Part 2: marginal IV fails

$$\mathrm{C}\mathrm{o}\mathrm{v}(Z,X)=\mathrm{V}\mathrm{a}\mathrm{r}(Z)=2.$$

$$\mathrm{C}\mathrm{o}\mathrm{v}(Z,Y)=2\mathrm{C}\mathrm{o}\mathrm{v}(Z,X)+2\mathrm{C}\mathrm{o}\mathrm{v}(Z,U)+3\mathrm{C}\mathrm{o}\mathrm{v}(Z,C)=2(2)+0+3(1)=7.$$

$${\beta }_{\mathrm{I}\mathrm{V}}=\frac{7}{2}=3.5.$$

This is biased. Relevance holds ($\mathrm{C}\mathrm{o}\mathrm{v}(Z,X)=2\ne 0$) and $Z⟂U$, but the exclusion/exogeneity condition fails: there is an open path $Z\leftarrow C\to Y$, so $Z$ is associated with $Y$ through a route that does not pass through $X$. The extra term is exactly $\kappa \lambda /(\alpha \mathrm{V}\mathrm{a}\mathrm{r}(Z))=3·1/(1·2)=1.5$ on top of $\beta$.

Part 3: conditional IV

Regress each variable on $C$ and take residuals. The projection coefficient of $Z$ on $C$ is $\mathrm{C}\mathrm{o}\mathrm{v}(Z,C)/\mathrm{V}\mathrm{a}\mathrm{r}(C)=1$, so

$$\tilde{Z}=Z-C={ϵ}_Z,\mathrm{V}\mathrm{a}\mathrm{r}(\tilde{Z})=1.$$

Because $\tilde{Z}={ϵ}_Z$ is orthogonal to $C$, partial covariances reduce to covariances with ${ϵ}_Z$:

$$\mathrm{C}\mathrm{o}\mathrm{v}(Z,X\mid C)=\mathrm{C}\mathrm{o}\mathrm{v}({ϵ}_Z,X)=\mathrm{C}\mathrm{o}\mathrm{v}({ϵ}_Z,Z+U+{ϵ}_X)=\mathrm{V}\mathrm{a}\mathrm{r}({ϵ}_Z)=1,$$

$$\mathrm{C}\mathrm{o}\mathrm{v}(Z,Y\mid C)=\mathrm{C}\mathrm{o}\mathrm{v}({ϵ}_Z,Y)=\mathrm{C}\mathrm{o}\mathrm{v}({ϵ}_Z,2X+2U+3C+{ϵ}_Y)=2\mathrm{C}\mathrm{o}\mathrm{v}({ϵ}_Z,X)=2.$$

$${\beta }_{\mathrm{c}\mathrm{I}\mathrm{V}}=\frac{2}{1}=2=\beta .$$

Conditioning on $C$ blocks the path $Z\leftarrow C\to Y$, restoring the exclusion condition, while relevance survives ($\mathrm{C}\mathrm{o}\mathrm{v}(Z,X\mid C)=1\ne 0$) and $Z⟂U\mid C$ still holds. Crucially, $U$ never had to be observed: an instrument only needs the confounder-to-outcome path from the instrument to be blocked, not the treatment's own confounding, which the instrument sidesteps rather than adjusts away.

Note

A tempting fix is ordinary backdoor adjustment: regress $Y$ on $X$ and $C$. This does not work, because the residual of $X$ after removing $C$ is $\tilde{X}={ϵ}_Z+U+{ϵ}_X$ (variance $3$), and $$\mathrm{C}\mathrm{o}\mathrm{v}(\tilde{X},Y)=2\mathrm{V}\mathrm{a}\mathrm{r}(\tilde{X})+2\mathrm{C}\mathrm{o}\mathrm{v}(\tilde{X},U)=2(3)+2(1)=8,$$ giving an $X$-coefficient of $8/3\approx 2.67$. No adjustment set built from observables blocks $X\leftarrow U\to Y$, since $U$ is unobserved — which is precisely why an instrument is needed here rather than backdoor adjustment. Final answers: ${\beta }_{\mathrm{O}\mathrm{L}\mathrm{S}}=3.25$, ${\beta }_{\mathrm{I}\mathrm{V}}=3.5$, ${\beta }_{\mathrm{c}\mathrm{I}\mathrm{V}}=2$.