1. Probability/functional equation/exponential distribution

(1) How would you solve the following: $\displaystyle g(s+t) = g(s)g(t)$?

(2) Suppose that the amount of time that a lightbulb works before burning itself out is exponentially distributed with mean ten hours. Suppose that a person enters the room in which the lightbulb is burning. If the person wants to work for $\displaystyle 5$ hours, what is the probability that he will be able to complete his work without the bulb burning out? What if its not exponential.

Ok so I think $\displaystyle P(\text{remaining lifetime} > 5 ) = e^{-1/2}$ for exponentially distribution. However, now I want to calculate $\displaystyle P(\text{lifetime} > t+5| \text{lifetime} > t)$.

I would think that this would equal $\displaystyle \frac{P(\text{lifetime} > t+5)}{P(\text{lifetime} > t)}$.

2. #1. If the function is continous see this.

3. well its right continuous.

4. Originally Posted by shilz222
[snip]
(2) Suppose that the amount of time that a lightbulb works before burning itself out is exponentially distributed with mean ten hours. Suppose that a person enters the room in which the lightbulb is burning. If the person wants to work for $\displaystyle 5$ hours, what is the probability that he will be able to complete his work without the bulb burning out? What if its not exponential.

Ok so I think $\displaystyle P(\text{remaining lifetime} > 5 ) = e^{-1/2}$ for exponentially distribution. Mr F says: Correct. See also my main reply.

However, now I want to calculate $\displaystyle P(\text{lifetime} > t+5| \text{lifetime} > t)$.

I would think that this would equal $\displaystyle \frac{P(\text{lifetime} > t+5)}{P(\text{lifetime} > t)}$. Mr F says: Correct.
The pdf is $\displaystyle \lambda e^{-\lambda t}$, t > 0, where $\displaystyle \mu = \frac{1}{\lambda} = 10$.

And as you might have discovered by now, the exponential distribution is 'memoryless':

$\displaystyle P(\text{lifetime} > t+5| \text{lifetime} > t) = \frac{P(\text{lifetime} > t+5)}{P(\text{lifetime} > t)} = P(\text{lifetime} > 5)$ for t > 0.

So now you need to calculate $\displaystyle \displaystyle \int_5^{+\infty} 0.1 e^{-0.1 t} \, dt$. This is equal to $\displaystyle e^{-1/2}$ .....

5. I thought $\displaystyle F(x) = P \{X \leq x \} = 1 - e^{-\lambda x}$ and thus the probability is $\displaystyle 1 - F(5)$. And it has to be right continuous.

6. Originally Posted by shilz222
I thought $\displaystyle F(x) = P \{X \leq x \} = 1 - e^{-\lambda x}$ and thus the probability is $\displaystyle 1 - F(5)$. And it has to be right continuous.
Yep, my mistake. I integrated for 0 < t < 5 by mistake. The reply's fixed now.

7. and then the second probability would be $\displaystyle \frac{1 - F(t+5)}{1- F(t)}$?

8. Originally Posted by shilz222
and then the second probability would be $\displaystyle \frac{1 - F(t+5)}{1- F(t)}$?
Yes. But did you read the bit in my post about memoryless ..... (I didn't screw up everything, you know )

9. well that probability is assuming that the distribution is NOT exponential.

10. Originally Posted by shilz222
well that probability is assuming that the distribution is NOT exponential.
I think it was clear from context that I was talking about the exponential distribution in all my posts.

When the distribution is not exponential, and you have no idea what the pdf is, then there's not much you can say except $\displaystyle \frac{1 - F(t+5)}{1- F(t)}$. But I don't see how that's really saying anything different to the (correct) probability statement that you gave ......

11. Yeah, but in the original question there were two parts. But thanks.

12. Originally Posted by shilz222
Yeah, but in the original question there were two parts. But thanks.
Aha. And thankyou for your thanks - lovely manners.