# What's the probability that you wait $2$ hours for the train when you have already waited $1$ hour?

Let's say there are independent waiting times of trains which are exponentially distributed with mean value $$\frac{1}{2}$$ hours. If you have already waited $$1$$ hour for the train, what's the probability that you will wait $$2$$ hours?

We have that mean value $$\mu = \frac{1}{2}$$. It's known that $$\mu = \frac{1}{\lambda} \Leftrightarrow \frac{1}{2} = \frac{1}{\lambda} \Leftrightarrow \lambda = 2$$

Now comes the part I'm not sure about.. when you have already waited $$1$$ hour before and now wait additional $$2$$ hours how the probability is calculated correctly here? I have looked this up on the internet and found that memorylessness applies to exponential distribution with formula:

$$P(X > r+t \mid X > r) = P(X>t)$$

where $$r$$ is the time you have previously waited, so $$r=1$$ and where $$t$$ is the time you have waited afterwards, thus $$t=2$$. Putting this into the formula we have

$$P(X>3 \mid X>1) = P(X>2) = 1-P(X<2) = 1-\left(1-e^{-2 \cdot2}\right) \approx 0.0183$$

So when you have already waited $$1$$ hour for the train, there is a probability of $$0.0183$$ that you will be waiting additional $$2$$ hours for the train?

Can you please tell me if it's correct like that because that's how I would do it in the exam :c

• The word "additional" is not in the problem statement. You have to find $P(X>2\mid X>1)=P(X>1)=e^{-2\cdot 1}$. – NCh Feb 3 at 1:17

Let $$T_n\stackrel{\mathrm{i.i.d.}}\sim\mathrm{Expo}(\lambda)$$. Define $$S_0=0$$ and $$S_n=\sum_{i=1}^n T_n$$. Then $$S_n$$ is a renewal process, in fact a Poisson process, with associated counting process $$N(t) = \sup\{n: S_n\leqslant t\}$$. Define the age process by $$A_t = t - S_{N(t)}$$ and the residual process by $$R_t = S_{N(t)+1} - t$$ for $$t\geqslant 0$$. Define the renewal function $$M(t) = \mathbb E[N(t)]$$. Since $$N(t)$$ is Poisson distributed with mean $$\lambda t$$, it readily follows that $$M(t)=\lambda t$$. Let $$Z_t\stackrel{\mathrm{def}}=(A_t,R_t)$$. We derive the distribution of $$Z_t$$ by considering when $$A_t=t$$, that is, when $$t$$ lies within the first renewal interval. Then we have $$f_{Z_t}(t,y) = f_{T_1}(t+y) = \lambda e^{-\lambda(t+y)}\cdot\mathsf 1_{[0,\infty)}(y).$$ The other case where $$A_t corresponds to when $$t$$ occurs after the first renewal. We have \begin{align} f_{Z_t}(x,y) &= \sum_{n=1}^\infty \frac{\lambda^n(t-x)^{n-1}e^{-\lambda(t-x)}}{(n-1)!}\lambda e^{-\lambda(x+y)}\ &= \lambda^2 e^{-\lambda(x+y)}\cdot\mathsf 1_{[0,t)\times[0,\infty)}(x,y).
$$f_{Z_t}(x,y) = \lambda^2e^{-\lambda(x+y)}\mathsf 1_{[0,t)\times[0,\infty)}(x,y) + \lambda^2e^{-\lambda(t+y)}\mathsf 1_{\{x=t\}\times[0,\infty)}(x,y)$$