Waiting for a train in a subway

[docnet]

Homework Statement

.

Relevant Equations

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(a) $$P(X≥10)=∫_{10}^∞10e^{−10x}dx=e^{−100}$$

(b) $$E(X)=∫_0^∞10xe^{−10x}dx=\frac{1}{10}$$

(c)$$\frac{10e^{−10}}{P(X≥10)}=10e^{100−10x}=\text{the new probability distribution for}\quad x≥10$$
$$E(X)=∫_{10}^∞10xe^{100−10x}dx=\frac{101}{10}$$

(d) imho it's unclear which "probability distribution functions" I should use here.
$$\begin{cases}
10e^{-10x} & 0\leq a\leq 10\\
0 & \text{otherwise}
\end{cases}$$
$$\begin{cases}
\frac{10e^{-10x} }{1-e^{-100}} & 0\leq a\leq 10\\
0 & \text{otherwise}
\end{cases}$$
The first doesn't satisfy the definition of the probability distribution function because its integral over $\mathbb{R}$ is less than 1. The second one has been normalized to 1 but it is not equal to $X\sim \exp⁡(10)$. The first leads to $\frac{1-101e^{-100}}{10}$ minutes. The second leads to $\frac{1}{10(1−e^{−100})}$ minutes which does not parse with the fact that train always arrives by 10 minutes at the latest, so it should be less than 110 minutes.

 

[PeroK]

First, at this level I would prefer to remain with the generic exponential distribution, as we can see more clearly what's going on. Just plug in $\lambda = 10$ and $t_0 = 10$ to get the numerical answers.

For example, for part a) we see that $$P(X \ge t_0) = e^{-\lambda t_0} = e^{-100}$$

For part c) I suggest you need to introduce a new variable $X' = X - t_0$, which represents the remaining time beyond $t_0$, and $X$ represents the total time including $t_0$.

I don't agree with your answer, because you want $E(X')$ instead of $E(X)$.

For part d), conceptually we have a reduced sample space of those cases where $X \le t_0$. We may, therefore, take the truncated distribution for $X \le t_0$ and scale it up by the relevant factor. In this case:
$$f(x) = \frac{\lambda e^{-\lambda x} } {\int_0^{t_0}\lambda e^{-\lambda x}dx} \ (0 \le x \le t_0)$$Which is equivalent to your second option.

 

[Office_Shredder]

Are you sure that by exp(10) they mean something that looks like $e^{-10 x}$ and not something like $e^{-x/10}$?

The question in general makes a lot more sense if it's the latter. Perhaps a typo in the book?

 

[docnet]

For part c) I suggest you need to introduce a new variable $X' = X - t_0$, which represents the remaining time beyond $t_0$, and $X$ represents the total time including $t_0$.

thank you :)

$$X'=X-t_0$$
$$E(X')=E(X)-E(t_0)=\frac{101}{10}-10=\frac{1}{10}$$
Do you think it strange the answers for (b) and (c) are same?

Are you sure that by exp(10) they mean something that looks like $e^{-10 x}$ and not something like $e^{-x/10}$?

The question in general makes a lot more sense if it's the latter. Perhaps a typo in the book?

yes, I'm sure the professor means $\lambda e^{-\lambda x}$

 

[PeroK]

Do you think it strange the answers for (b) and (c) are same?

No, not strange. Inevitable, if you do the maths. Once an exponential, always an exponential!

 

[BvU]

And nobody is objecting ? If the waiting time is really six seconds, the train can't even stop !

$\$

 

[PeroK]

And nobody is objecting ? If the waiting time is really six seconds, the train can't even stop !

$\$

It's called a bullet train!

 

[PeroK]

No, not strange. Inevitable, if you do the maths. Once an exponential, always an exponential!

To see this, consider the distribution for $x' = x - t_0$ in the cases where we have waited a time $t_0$ without a train. We have:
$$P(X \ge t_0) = \int_{t_0}^{\infty} \lambda e^{-\lambda x} dx = e^{-\lambda t_0}$$That's the size of the new sample space. And:
$$f(x') = \frac{\lambda e^{-\lambda(x' + t_0)}}{e^{-\lambda t_0}} = \lambda e^{-\lambda x'}$$Which shows that the distribution for the remaining time remains the same exponential distribution.

That's a good example of what you discover when you stick to the general case and resist the temptation to plug and chug!