Stochastic modeling. A bidding Model The Question: 
Let $U_1,U_2,...$ be independent RVs, each uniformly distributed over $(0,1]$. These random variables represent successive bids on an asset that you're trying to sell, and that you must sell by time $t = 1$. As a strategy you adopt a secret number $θ$, and you will accept the first offer that is greater than $θ$. For example, you accept the second offer if $U_1≤θ$ while $U_2>θ$. Suppose that the offers arrive according to a unit rate Poisson process ($λ=1$). 


*

*What is the probability that you sell the asset by time $t=1$? 

*What is the value for $θ$ that maximizes your expected return? (You get nothing if you don't sell the asset by $t = 1$).

*To improve your return you adopt a new strategy, which is to accept an offer at time $t$ if it exceeds $θ(t)=(1−t)/(3−t)$. What are your new chances of selling the asset, and what is your new expected return?


Just for the first part (1) I'm still quite unsure what to do. What seems somewhat reasonable to me is to calculate $$Pr[X_T(t)>θ|X(t)>0]$$ where $T=\max(U_1,...,U_T>θ)$.  I'm not sure how to calculate this probability either though. Hints and clarifications would be very much appreciated thanks.
 A: Let $\{N(t):t\geqslant 0\}$ be the arrival process with arrival times $\{T_n\}$ and $$M(t) = \sum_{n=1}^{N(t)} T_n\mathsf 1_{(\theta,1]}(U_n). $$ Then $\{M(t):t\geqslant 0\}$ is a Poisson process with rate $1-\theta$, and the probability you sell the asset by time $t=1$ is $$\mathbb P(M(1)>0) = 1-\mathbb P(M(1)=0) = 1 - e^{-(1-\theta)}. $$
The expected return given threshold $\theta$ is $$\mathbb E[U_1\mid U_1>\theta]\mathbb P(M(1)>0) =\left(\frac{1+\theta}2\right)\left(1-e^{-(1-\theta)}\right),$$ which is maximized at $$\theta = W(e^3)-2\approx0.20794$$ where $W$ is the Lambert $W$-function.
For part 3, $M(t)$ is a nonhomogenous Poisson process with intensity function $1-\theta(t)$. Let $$\Theta(t) = \int_0^t (1-\theta(s))\ \mathsf ds$$ for $0<t\leqslant 1$, then the expected number of arrivals in $(0,t]$ is $$\Theta(t) = \int_0^t \left(\frac2{3-t}\right)\ \mathsf ds = 2(\log(3) -\log(3-t)) $$
In particular, $\Theta(1)=\log\left(\frac 94\right)$, so the probability of selling is $$1-\mathbb P(M(t)=0) = 1-e^{-\Theta(1)}=\frac59. $$
Conditioned on $\{T_1=t\}$, the expected return is $\frac{1+\theta(t)}2=\frac{2-t}{3-t}$. So the expected return is given by $$\int_0^1 \left(\frac{2-t}{3-t}\right)f_{T_1}(t)\ \mathsf dt = \int_0^1 \frac{2(2-t)e^{-2(\log(3)-\log(3-t))}}{(3-t)^2}\ \mathsf dt =\frac13. $$
