On the expectation of normalized Brownian functionals up to first hitting times

Let B be a Brownian motion and T its first hitting time of the level 1. For U a uniform random variable independent of B, we study in depth the distribution of T^{-1/2}B_{UT}, that is the rescaled Brownian motion sampled at uniform time. In particular, we show that this variable is centered.


Introduction
In this paper, we study the expectations of the random variables A where B is a Brownian motion and T a denotes the first hitting time of the level a by B. First, remark that T a /a 2 is the first hitting time of a by (B a 2 s ).Therefore, the scaling property of the Brownian motion implies that the laws of A (m) a and Ã(m) a do not depend on a.
To fix ideas, let us now focus in this introduction on the variables A (m) a .These variables are clearly asymmetric functionals of the Brownian motion.Nevertheless, we may wonder whether there exist values of m such that A (m) a is centered (we will show later that these variables have moments of all orders).Indeed, consider for example the case where m is an odd integer: using a symmetry argument, it is clear that where −a is obviously defined.Since these two quantities do not depend on a, we get a given value, say v m , for the expectations when the barrier is positive and −v m when it is negative.This somewhat suggests that v m may be equal to zero.
In fact, it turns out that the random variable A (m) a is centered only for m = 1.This result has several interesting consequences.In particular, we show that it can be very simply interpreted in terms of the Brownian meander.Moreover, we prove that the expectation of A (m) a is negative for m < 1 and positive for m > 1.
Finally, note that these expectations are closely connected with the random variable α defined by where U is a uniform random variable, independent of B. For example, for m an odd integer, the m-th moment of α is the expectation of A (m) 1 .This led us to give in Theorem 3.2 the law of α.
The paper is organized as follows.The specific case m = 1 is treated in Section 2. Our main theorem which provides the expectations of A Theorem 2.1 states that, as far as the expectation is concerned, between 0 and T 1 , the time spent by the Brownian motion in (−∞, 0) is balanced by that spent in [0, 1].Again, it is tempting to deduce this result from the scaling and symmetry properties of A (1) 1 .However, Theorem 3.1 will formalize that such intuition is wrong.Indeed, we will for example show that the expectation of A (3) 1 is non zero, although it satisfies the same scaling and symmetry properties as A (1) 1 .In fact, we will see that the expectation of A (m) 1 is strictly positive for m > 1 and stricly negative for m < 1.
Theorem 2.1 can in fact be interpreted as a corollary of the general result given in Theorem 3.1 below.However, using Williams time reversal theorem and some absolute continuity results for Bessel processes, a specific, elegant proof can be written for Theorem 2.1.So we give this proof in Appendix A.

More integrability properties for A
(1) 1

and connection with Knight's identity
Let (L t ) t≥0 be the local time process at 0 of the Brownian motion B and set for l > 0. Recall that Lévy's equivalence result, gives the following equality: with S t = sup s≤t B s and = L denotes equality in law, see [10].Thus, we obtain that A (1) 1 has the same law as the random variable ζ defined by We obviously have On the one hand, it is well known that 1/ √ τ 1 follows the law of the absolute value of a standard Gaussian random variable.On the other hand, the celebrated Knight's identity states the following equality: where T 3 2 = inf{t, R t = 2}, with R a three dimensional Bessel process, see [7].Using the scaling property of the three dimensional Bessel process, we easily get the equality: Therefore, we deduce that 1 √ τ 1 sup Hence, we easily deduce the following proposition: Proposition 2.1.There exists ε > 0 such that We note that the same arguments yield that A (m) 1 and Ã(m) admit moments of all orders.

Consequences of Theorem 2.1 for the Bessel process, Brownian meander and Brownian bridge
We give in this subsection some corollaries of Theorem 2.1 involving very classical processes, namely the three dimensional Bessel process, the Brownian meander, and the Brownian bridge.We start with a result about the three dimensional process, whose proof is given within the proof of Theorem 2.1 in Appendix A.
Corollary 2.1.Let (R t ) t≥0 denote a three dimensional Bessel process.We have Thanks to Imhof's relation, see [3,6], we immediately deduce the following result from Corollary 2.1: Corollary 2.2.Let (m t ) t≤1 be the Brownian meander.We have We now give a corollary involving the Brownian bridge.
Corollary 2.3.Let (b t ) t≤1 denote the Brownian bridge and (l t ) t≤1 its local time at zero.We have Proof.From [4], we get the following equality: Thus, using Corollary 2.2, we get Now remark that the process ( bt ) = (b 1−t ) is also a Brownian bridge whose local time at time t, denoted by lt , satisfies lt = l 1 − l 1−t . Consequently, This implies and therefore Equation ( 1) provides Finally, using a pathwise transformation between the meander and the Brownian excursion, see [2], Corollary 2.2 also enables to show the following result: Corollary 2.4.Let (e t ) t≤1 denote the standard Brownian excursion.We have

The case of two barriers
After the striking result given in Theorem 2.1, it is natural to wonder whether the expectation remains equal to zero if T a is replaced by T a,b , where T a,b is the first exit time of the interval (−b, a), with a > 0 and b > 0. Indeed, remark that the random variable A a,b defined by still enjoys a scaling property in the sense that its law only depends on the ratio b/a.In fact, the following theorem states that the expectation is no longer zero in this case1 : Theorem 2.2.Let λ = b/a.We have The proof of this result is given in Appendix B. In fact a general formula for with θ > 0 is given within this proof.Eventually, note that Theorem 2.1 can also be recovered from Theorem 2.2 letting the downward barrier tend to −∞.
3 The general case

Computation of the expectations
For x ∈ R, we set x + = max(x, 0) and x − = max(−x, 0).For m ≥ 0, we define with the convention 0 0 = 0. We also write and Furthermore, we note that I (m) ± is the moment of order m of the random variable α ± where with U a uniform random variable independent of the Brownian motion B. We study the variable α in more details in Section 3.3.
For m ≥ 0, let where N is a standard Gaussian random variable and Γ denotes the Gamma function.We have the following theorem: The following formulas hold: In particular, we note that φ(0 (3).We give the proof of Theorem 3.1 in Section 3.6.

Comments about Theorem 3.1
• The function φ is well defined for m ∈ (−2, +∞) and satisfies φ(−1) = φ(1) = log(3).Thus, we retrieve in Theorem 3.1 the fact that E[A • We easily get that φ is twice differentiable and, for m ≥ 0, Hence φ is convex and furthermore, we show in Appendix C that φ (0) > 0. This implies that φ and φ are increasing on R + .Hence, since we get I (m) > 0 for m > 1 and I (m) < 0 for m < 1.This can be interpreted as follows: from the point of view of Within the proof of Theorem 3.1, we are led to show the following interesting result: and We also give another proof of Proposition 3.1, based on the Ray-Knight theorem, in Appendix D.

Uniform sampling up to hitting time
We now want to interpret Theorem 3.1 as a result about sampling independently and uniformly the properly rescaled Brownian motion up to its first hitting time T 1 .More precisely, let us introduce (l y 1 , y ∈ R), the local time at time 1 of the process Let f be a Borel non negative function and U a uniform random variable independent of any other random variable defined here.Using the occupation formula, we get is the density of α at point y.The following result is easily deduced from Theorem 3.1, by injectivity of the Mellin transform.
Theorem 3.2.The density h satisfies for y ≥ 0 and for y ≤ 0 Hence, conditional on α > 0, the law of α + is a mixture of absolute Gaussian laws, whereas conditional on α < 0, α − is distributed as the absolute value of a Gaussian random variable.
Remark that for y ≥ 0, we have the obvious inequality Therefore, we have the following corollary: In fact, thanks to Proposition 3.1, we can even provide the density at point y of α conditional on T 1 = t.We denote this density by h(y, t).Obvious relations between (l y 1 ) and (L y t ) yield We easily obtain the following corollary from Proposition 3.1: and for x ≥ 0

Interpretation in terms of the Brownian meander
In the same spirit as in Corollary 2.2, we can give an interpretation of Theorem 3.1 in terms of the Brownian meander.Using Williams theorem in the same way as in the proof of Theorem 2.1, see Appendix A, together with Imhof's relation, see [3,6], as already done for Corollary 2.2, we get that for any non negative measurable functions f and g, where m denotes the Brownian meander.Let U be a uniform random variable, independent of all other quantities.The last relation is equivalent to Thus, from Corollary 3.2, we are able to compute the density of m U conditional on the value m 1 .More precisely, we have the following theorem: Let f be a Borel non negative function.We have

Future developments
In this work, we have studied some properties of random sampling through the random variable Another interesting variable is the variable β defined by with τ l = inf{t ≥ 0, L t > l}.In fact the associated process is called pseudo Brownian bridge and has been considered more explicitly in the literature than In particular, it enjoys some absolute continuity property with respect to the standard Brownian bridge, see [3].We intend to present results related to β in a forthcoming work, in a way which will help us to recover the interesting law of α.For now, we only mention that β is distributed as N/2, where N is a standard Gaussian random variable.

Proof of Theorem 3.1
Let m ≥ 0. We split the proof into several steps.
Step 1: Introducing a natural measure First, let us remark that Hence, it is natural to introduce for µ ≥ 0 the measure I µ , which to a positive function ψ associates Step 2: Computation of I µ (ψ) Using the martingale property of the process exp(µB s − µ 2 s/2), we get We now use the following well known formula, see for example [9]: for s > 0 and b ∈ R, It implies that I µ (ψ) is equal to which can be rewritten Then, using the density and the value of the first moment of an inverse Gaussian random variable, we get that for µ > 0 and y ∈ R, From this, we deduce that when the support of ψ is included in [0, 1], and when the support of ψ is included in (−∞, 0), Remark here that Proposition 3.1 immediately follows from Equation (2) and Equation (3).
Step 3: End of the proof of Theorem 3.1 We end the proof of Theorem 3.1 in this final step.We start with the following elementary lemma: Applying Lemma 3.1, we obtain the result for I

Appendices
A Proof of Theorem 2.1 Theorem 2.1 can be seen as a particular case of Theorem 3.1.Nevertheless, we give here a specific proof for this theorem which is interesting on its own.We split it in several steps.
Step 1: Time reversal Let us recall Williams time reversal theorem, see for example [10].We have the following equality: (1 where R denotes a three dimensional Bessel process starting from 0 and γ is its last passage time at level 1: Consequently, since it has the same law as Step 2: Moments We now show that A 1 has moments of any order.First recall the following equalities: Thus, 1 √ γ has moments of any order and therefore it is enough to prove the integrability of ξ r , for any r > 0, with Such integrability result will be deduced from the following absolute continuity relation that can be found in [3]: Now take r > 0, 1 < p < 3/2 and q such that 1/p + 1/q = 1.From Lemma A.1 together with Hölder inequality, we obtain The first expectation on the right hand side of the last inequality is obviously finite.For the second one, recall that R 2 1 has the distribution of 2Z, with Z following a gamma law with parameter 3/2.Therefore, the second expectation is also finite since p < 3/2.

Step 3: Centering property
We end the proof of Theorem 2.1 in this step.We start with the following technical lemma.
Lemma A.2. Let a > 0. We have Proof.Using again that R 2  1 has the distribution of 2Z, with Z following a gamma law with parameter 3/2, we can write The result follows easily from this equality.

We now prove that E[A (1)
1 ] = 0. From Equation (4) and Lemma A.1, using the fact that this is equivalent to prove the following lemma: Lemma A.3.We have Proof.First, using Markov property, we get where E r denotes the expectation of a three dimensional Bessel process starting from point r.From Proposition 2, page 99, in [11], we know that Thus, using the last equality together with a change of variable and the scaling property of the Bessel process, we get From Lemma A.2, we obtain Using Fubini's theorem when integrating in u from 0 to 1, and remarking that Γ(3/2) = √ π/2, we easily conclude the proof of Lemma A.3 and so the proof of Theorem 2.1.

B Proof of Theorem 2.2
Let a > 0, b > 0 and θ > 0. In this section, we consider ψ(a, b, θ) defined by where τ is the exit time of the interval (−b, a) by the Brownian motion B.

B.1 General result
We start with a general result.We give here a representation of ψ(a, b, θ) in term of a Lebesgue integral.Let δ > 0, a > 0, b > 0 and p > −1.Recall that c p denotes the p−th absolute moment of a standard Gaussian random variable and define φ δ (a, b, p) by We have the following result.
Theorem B.1.Let θ > 0. We have Proof.Our proof is based on Feynman-Kac formula, see for example [5].Note that in [8], the author used this formula in order to derive the joint Laplace transform of (τ, τ 0 B s ds).We propose here a specific method for our problem.We introduce the function g : (x, δ, ρ) → E x e −(δ 2 /2)τ +ρ τ 0 Bsds .
By Feynman-Kac formula, g solves on (−b, a) For ρ = 0, we denote g 0 : (x, δ) → g(x, δ, 0) which solves on (−b, a) Thus, g 0 is of the form Differentiating the dynamics of g with respect to ρ and introducing we observe that f solves on (−b, a) Furthermore, by definition of g, f satisfies Due to its dynamics, we get that f is of the form where f 0 is a particular solution of the ODE of interest.Applying the variation of the constant method, we look for f 0 of the form This function f is of particular interest since for p > −1, Hence, denoting p = 2θ − 1, the first part of Theorem B.1 boils down to the computation of so that we need to identify A δ , B δ , C δ (.), D δ (.), E δ and F δ .
We deduce that A δ and B δ are given by : Similarly we can compute E δ and F δ in terms of C δ (.) and D δ (.).Indeed, the boundary conditions of f imply Consequently, we get that E δ sh(δ(a + b)) is equal to It now remains to compute C δ (.) and D δ (.), which are both defined up to a constant and satisfy Therefore, we get We now compute C δ , which is given by In the same way, D δ is given by Since C δ (0) = 0, observe that the quantity of interest (5) rewrites Plugging the values for the coefficients, this can be rewritten which leads to the expression: After obvious computations, we obtain that it is also equal to By definition, A δ ch(δa) + B δ sh(δa) = A δ ch(δb) − B δ sh(δb) = 1.Therefore, we get Recall also that A δ and B δ are explicitly given by (6) so that A δ sh(δa) + B δ ch(δa) = sh(δb)sh(δa) + sh(δa) 2 + ch(δb)ch(δa) − ch(δa) Plugging these expressions in the previous one provides Recalling that p = 2θ − 1, this ends the proof of the first part of Theorem B.1.
We now give the proof of the second part.By integration by parts we get that Then, we easily obtain that the last expression is equal to After obvious simplifications, this can be rewritten Thus, using the function φ δ defined before Theorem B.1, we obtain

B.2 Proof of Theorem 2.2
We now give the proof of Theorem 2.2.From Theorem B.1, we get We finally obtain the result after the change of variable x = δa.
C Some computations about the function φ defined in Theorem 3.1 Recall that the function φ is defined for m > −2 by φ(m) = We have Proof.We get the equality of the two functions in Lemma C.1 by showing that they have the same derivatives and that they coincide for C = 1.To show the equality of the derivatives, after straightforward computations, we see that we need to prove that is equal to zero.Now we use the fact that for |x| ≤ 1, see [1], in order to get the result.
We now show that the values of the two functions in Lemma C.1 coincide for C = 1.We have We conclude using the fact that Li 2 (−1) = −π 2 /12, see again [1].
Eventually, we give the graphs of the functions φ, φ and ∆ in Figures 1, 2 and 3.This ordinary differential equation can be easily solved using the variation of the constant method so that we get The proof for L −x T 1 (µ) goes similarly.

x n n 2 ,
We denote by Li 2 the dilogarithm function defined for x such that |x| ≤ 1 by Li 2 (x) = +∞ n=1 see[1] for more details.We start with the following general lemma: Lemma C.1.For C ≥ 1, we define the function ∆ by