Wednesday 26 August 2026

Part 1: reflection principle and the joint density

Fix $a\ge 0$ and $b\le a$. On the event $\{M_T\ge a\}$ the stopping time ${\tau }_a=inf\{t:W_t=a\}$ satisfies ${\tau }_a\le T$. By the strong Markov property, the post-${\tau }_a$ process ${\tilde{W}}_s=W_{{\tau }_a+s}-a$ is a standard Brownian motion independent of ${\mathcal{F}}_{{\tau }_a}$, and by symmetry $-\tilde{W}$ has the same law. Reflecting the path after ${\tau }_a$ about the level $a$ therefore produces an equally likely path, and it sends the endpoint $W_T=w$ to $2a-w$.

Under this reflection the event $\{M_T\ge a,W_T\le b\}$ (endpoint at least $a-b$ below the barrier) maps bijectively onto $\{M_T\ge a,W_T\ge 2a-b\}$. But since $b\le a$ we have $2a-b\ge a$, so $\{W_T\ge 2a-b\}\subseteq \{M_T\ge a\}$ automatically. Hence $$\mathbb{P}(M_T\ge a,W_T\le b)=\mathbb{P}(M_T\ge a,W_T\ge 2a-b)=\mathbb{P}(W_T\ge 2a-b).$$

The key step a candidate misses is checking $2a-b\ge a$, which is exactly what lets us drop the constraint $M_T\ge a$ on the right.

Now extract the density. Write ${\varphi }_T(x)=\frac{1}{\sqrt{2\pi T}}{e}^{-x^2/2T}$. Differentiating $$\mathbb{P}(M_T\ge a,W_T\le b)={\int }_{2a-b}^{\infty }{\varphi }_T(x)dx$$ in $b$ gives ${\int }_a^{\infty }g(m,b)dm={\varphi }_T(2a-b)$ (chain rule contributes $+1$ from ${\partial }_b(-(2a-b))$). Differentiating again in $a$: $$-g(a,b)={\partial }_a{\varphi }_T(2a-b)=-\frac{2(2a-b)}{T}{\varphi }_T(2a-b).$$ Therefore the joint density is $$g(m,w)=\frac{2(2m-w)}{T\sqrt{2\pi T}}\exp (-\frac{(2m-w{)}^2}{2T}),m\ge 0,w\le m.$$

Part 2: law of $M_T$ and its mean

Setting $b=a$ in part 1 gives $\mathbb{P}(M_T\ge a,W_T\le a)=\mathbb{P}(W_T\ge a)$. Adding $\mathbb{P}(M_T\ge a,W_T>a)=\mathbb{P}(W_T>a)$ (since $W_T>a$ forces $M_T\ge a$), $$\mathbb{P}(M_T\ge a)=2\mathbb{P}(W_T\ge a)=\mathbb{P}(|W_T|\ge a).$$ So $M_T\stackrel{d}{=}|W_T|$, the half-normal law. Since $W_T~N(0,T)$, $$𝔼[M_T]=𝔼[|W_T|]=\sqrt{T}𝔼[|Z|]=\sqrt{T}·\sqrt{\frac{2}{\pi }}=\sqrt{\frac{2T}{\pi }}.$$

Part 3: adding drift via Girsanov

The reflection principle relies on the symmetry of Brownian increments and fails once there is drift; one cannot simply write $2\mathbb{P}(X_T\ge a)$. The clean route is a change of measure.

Define $\mathbb{Q}$ on ${\mathcal{F}}_T$ by $$\frac{d\mathbb{P}}{d\mathbb{Q}}{|}_{{\mathcal{F}}_T}=\exp (\mu X_T-\frac{1}{2}{\mu }^{2}T).$$ By Girsanov's theorem, under $\mathbb{Q}$ the process $X_t$ is a standard Brownian motion (the density removes the drift). Writing $M_T^X={max}_{t\le T}{X}_t$, $$\mathbb{P}(M_T^X\ge a)={𝔼}^{\mathbb{Q}}\left[{\mathbf{1}}_{\{M_T^X\ge a\}}{e}^{\mu X_T-\frac{1}{2}{\mu }^{2}T}\right],$$ and under $\mathbb{Q}$ the pair $(M_T^X,X_T)$ has the density $g$ from part 1. Since $a>0$, the constraint $m\ge a$ subsumes $m\ge 0$: $$\mathbb{P}(M_T^X\ge a)=e^{-{\mu }^{2}T/2}{\int }_{-\infty }^{\infty }{e}^{\mu w}{\int }_{max(a,w)}^{\infty }\frac{2(2m-w)}{T\sqrt{2\pi T}}{e}^{-(2m-w{)}^2/2T}dmdw.$$ The inner integral, with $u=2m-w$, is ${\varphi }_T(max(2a-w,w))$. Splitting at $w=a$ (where $2a-w=w$): $$\mathbb{P}(M_T^X\ge a)=e^{-{\mu }^{2}T/2}[{\int }_a^{\infty }{e}^{\mu w}{\varphi }_T(w)dw+{\int }_{-\infty }^a{e}^{\mu w}{\varphi }_T(2a-w)dw].$$ Completing the square in the first integral, $\mu w-\frac{w^2}{2T}-\frac{{\mu }^{2}T}{2}=-\frac{(w-\mu T{)}^2}{2T}$, gives $\Phi \left(\frac{\mu T-a}{\sqrt{T}}\right)$. In the second, substitute $v=2a-w$; the exponent becomes $-\frac{(v+\mu T{)}^2}{2T}$ times $e^{2\mu a}$, giving $e^{2\mu a}\Phi \left(\frac{-\mu T-a}{\sqrt{T}}\right)$. Hence $$\boxed{\mathbb{P}({max}_{0\le t\le T}{X}_t\ge a)=\Phi \left(\frac{\mu T-a}{\sqrt{T}}\right)+e^{2\mu a}\Phi \left(\frac{-\mu T-a}{\sqrt{T}}\right).}$$

Sanity check. At $\mu =0$ this collapses to $2\Phi (-a/\sqrt{T})=2\mathbb{P}(W_T\ge a)$, matching part 2.

Numerical value for $\mu =1,a=2,T=1$: $$\Phi (-1)+e^4\Phi (-3)=0.15866+54.598\times 0.0013499\approx 0.15866+0.07370=0.2324.$$

So the drifted Brownian motion reaches level $2$ within one time unit with probability $\approx 0.232$.

Closing note. The tempting wrong answer in part 3 is $2\mathbb{P}(X_T\ge a)=2\Phi \left(\frac{\mu T-a}{\sqrt{T}}\right)$, obtained by blindly reusing the driftless reflection identity. The $e^{2\mu a}$ factor is precisely the correction the change of measure supplies; forgetting it (or getting the sign of $\mu$ wrong inside the second $\Phi$) is the standard way to lose the problem.

1. Conjugacy

The Poisson likelihood is $p(x_{1:n}\mid \lambda )={\prod }_i\frac{{\lambda }^{x_i}{e}^{-\lambda }}{x_i!}\propto {\lambda }^S{e}^{-n\lambda }$ (as a function of $\lambda$; the $x_i!$ are constants). Multiplying by the prior kernel,

$$\pi (\lambda \mid x_{1:n})\propto {\lambda }^{\alpha -1}{e}^{-\beta \lambda }·{\lambda }^S{e}^{-n\lambda }={\lambda }^{(\alpha +S)-1}{e}^{-(\beta +n)\lambda }.$$

This is the kernel of a Gamma density, and since the posterior is a genuine probability density the normalizing constant is forced. Hence

$$\lambda \mid X_{1:n}~\mathrm{G}\mathrm{a}\mathrm{m}\mathrm{m}\mathrm{a}(\alpha +S,\beta +n).$$

The key structural fact is that the Poisson likelihood, as an exponential family in $\lambda$, has sufficient statistic $(S,n)$; the Gamma prior is the conjugate family whose hyperparameters live in the same coordinates, so updating is just addition of $(S,n)$ to $(\alpha ,\beta )$.

2. Posterior mean as shrinkage

For $\mathrm{G}\mathrm{a}\mathrm{m}\mathrm{m}\mathrm{a}(a,b)$ the mean is $a/b$, so

$$𝔼[\lambda \mid X_{1:n}]=\frac{\alpha +S}{\beta +n}.$$

Split the numerator to expose the two estimators:

$$\frac{\alpha +S}{\beta +n}=\frac{\beta }{\beta +n}·\frac{\alpha }{\beta }+\frac{n}{\beta +n}·\frac{S}{n}.$$

So the posterior mean is a convex combination

$$𝔼[\lambda \mid X_{1:n}]=(1-w)\underset{\text{prior mean}}{\underbrace{\frac{\alpha }{\beta }}}+w\underset{\text{MLE}}{\underbrace{\overline{X}}},w=\frac{n}{\beta +n}.$$

The weight on the data is $w=n/(\beta +n)\to 1$ as $n\to \infty$: the prior washes out at rate $O(1/n)$. The step a candidate skips is reading off what $\beta$ is: it enters exactly where $n$ does, so $\beta$ is a prior sample size — a count of pseudo-observations (equivalently, prior units of exposure), with $\alpha$ the pseudo-total of events. The prior $\mathrm{G}\mathrm{a}\mathrm{m}\mathrm{m}\mathrm{a}(\alpha ,\beta )$ is worth $\beta$ observations carrying $\alpha$ events.

3. Posterior predictive

Write ${\alpha }^{\prime }=\alpha +S$, ${\beta }^{\prime }=\beta +n$ for the posterior parameters. The predictive is the Poisson likelihood averaged against the posterior:

$$P(X_{n+1}=k\mid X_{1:n})={\int }_0^{\infty }\frac{{\lambda }^k{e}^{-\lambda }}{k!}·\frac{{\beta }^{\prime {\alpha }^{\prime }}}{\Gamma ({\alpha }^{\prime })}{\lambda }^{{\alpha }^{\prime }-1}{e}^{-{\beta }^{\prime }\lambda }d\lambda .$$

Collect the powers and use ${\int }_0^{\infty }{\lambda }^{c-1}{e}^{-d\lambda }d\lambda =\Gamma (c)/d^c$ with $c=k+{\alpha }^{\prime }$, $d={\beta }^{\prime }+1$:

$$P(X_{n+1}=k\mid X_{1:n})=\frac{{\beta }^{\prime {\alpha }^{\prime }}}{k!\Gamma ({\alpha }^{\prime })}·\frac{\Gamma (k+{\alpha }^{\prime })}{({\beta }^{\prime }+1{)}^{k+{\alpha }^{\prime }}}.$$

Rearrange:

$$P(X_{n+1}=k\mid X_{1:n})=\frac{\Gamma (k+{\alpha }^{\prime })}{k!\Gamma ({\alpha }^{\prime })}{\left(\frac{{\beta }^{\prime }}{{\beta }^{\prime }+1}\right)}^{{\alpha }^{\prime }}{\left(\frac{1}{{\beta }^{\prime }+1}\right)}^k,k=0,1,2,\dots$$

This is a negative binomial law with (real-valued) size $r={\alpha }^{\prime }=\alpha +S$ and success probability $p={\displaystyle \frac{{\beta }^{\prime }}{{\beta }^{\prime }+1}}={\displaystyle \frac{\beta +n}{\beta +n+1}}$:

$$X_{n+1}\mid X_{1:n}~\mathrm{N}\mathrm{e}\mathrm{g}\mathrm{B}\mathrm{i}\mathrm{n}(r=\alpha +S,p=\frac{\beta +n}{\beta +n+1}).$$

Its mean is $r(1-p)/p={\alpha }^{\prime }/{\beta }^{\prime }=𝔼[\lambda \mid X_{1:n}]$, and its variance $r(1-p)/p^2=\frac{{\alpha }^{\prime }}{{\beta }^{\prime }}·\frac{{\beta }^{\prime }+1}{{\beta }^{\prime }}>\frac{{\alpha }^{\prime }}{{\beta }^{\prime }}$ exceeds the mean: the predictive is overdispersed relative to a plug-in Poisson, because it carries the posterior uncertainty in $\lambda$ that a plug-in throws away.

Closing note

A quick coherence check that also gives a slicker route: because updating just adds $(S,n)$ to $(\alpha ,\beta )$, processing the data one point at a time and processing all $n$ at once yield the same posterior — Bayesian updating is associative here precisely because the sufficient statistic is additive. The tempting error in part 3 is to "plug in" $\widehat{\lambda }=\overline{X}$ and report $\mathrm{P}\mathrm{o}\mathrm{i}\mathrm{s}\mathrm{s}\mathrm{o}\mathrm{n}(\overline{X})$; that discards the integration over $\lambda$ and understates predictive variance.

Answers. (1) $\mathrm{G}\mathrm{a}\mathrm{m}\mathrm{m}\mathrm{a}(\alpha +S,\beta +n)$. (2) $𝔼[\lambda \mid X_{1:n}]=\frac{\beta }{\beta +n}\frac{\alpha }{\beta }+\frac{n}{\beta +n}\overline{X}$, weight $w=n/(\beta +n)\to 1$; $\beta$ is a prior sample size. (3) $\mathrm{N}\mathrm{e}\mathrm{g}\mathrm{B}\mathrm{i}\mathrm{n}(r=\alpha +S,p=\frac{\beta +n}{\beta +n+1})$ with the pmf displayed above.