Conditional expectation of a Poisson process with a given parameter Let $(N_{t})_{t\geq 0}$  be a Poisson process with $\lambda =2$. Let $W_{n}$ denote the time until the $n$th event.  The goal of this problem is to compute 
$$\mathbb{E}(W_{5}\mid N_{3}=5).$$
My first thought was to think of each $N_{t}$ as being uniformly distributed on the interval $(0,3).%$ Using this logic, the expectation above is equal to 
$$\mathbb{E}(\max(U_{1},U_{2},...,U_{5})),$$ 
with each $U _{i}$ being uniformly distributed on the interval $(0,3).$ 
The c.d.f. of $\max(U_{1},U_{2},...,U_{5})$, let's call it $F_{M}(t)$, is 
$$\mathbb{P}(M< t) = \prod_{i=1}^{5}\mathbb{P}(U_{i}< t) = (\frac{t}{3})^{5}.$$
The p.d.f. is then 
$$f_{M}(t)= \frac{\mathrm{d} }{\mathrm{d} t}F_{m}(t)= \frac{5}{3}(\frac{t}{3})^{4}.$$
Finally, we can find the expected value.
$$\mathbb{E}(W_{5}\mid N_{3}=5) = \int tf_{M}(t)dt = \frac{5}{3^5}\int_{0}^{3}t^5dt= \frac{5}{2}.$$
The logic seems correct; however, I never take in account the parameter $\lambda.$  If my answer is correct, why does the expected value not depend on the parameter of the process?  If my answer is not correct, what would be the correct way to approach this? 
Thank you in advance.   
 A: Your answer is correct. You could have obtained it without the detour through the density by calculating
$$
\mathbb{E}(W_{5}\mid N_{3}=5)=\int_0^3\mathbb P(M\ge t)\,\mathrm dt=\int_0^3\left(1-\left(\frac t3\right)^5\right)\,\mathrm dt=3-\frac36=\frac52\;.
$$
Or you could have obtained it without any calculus at all: Start with a circle, uniformly randomly select $6$ points on it; then uniformly randomly select one of them to be the point at which you cut the circle to produce the interval. This symmetric construction shows that, contrary to how it might appear at first sight, the $6$ intervals marked by the $5$ points actually all have the same distribution. Thus they all have the same mean $\frac12$, and thus the expected position of the $m$-th point is $\frac m2$.
The result doesn't depend on the parameter $\lambda$ because of the translation invariance of the Poisson process. The events are equally likely to occur anywhere in the interval, no matter with what rate they occur. Don't think of it as a process where you move from $0$ to $3$ through time; think of $[0,3]$ as a segment on which raindrops are falling, and watch until $5$ raindrops have fallen. It doesn't matter what rate they were falling at; once $5$ have fallen, the expected position of the right-most one is $\frac52$.
