# Finding A. Prob.

1. Oct 22, 2013

### Jbreezy

1. The problem statement, all variables and given/known data

The expoential distribution is given by

f(x) = 0 for x <0 and f(x) = Ae^-cx for x >= 0

2. Relevant equations

3. The attempt at a solution

Find A so that f(x) is a probability distribution. So all that I did was use the def.

I said A∫e^(-cx) dx = 1 from 0 to ∞ I integrated it I got that A/c = 1 so A = c.

So then the next question says find P(x <1) so I did ∫ce^-ct dt = -e^(-ct) between [0,1] so I'm just confused because I have c still it isn't like it's going to drop out. So I'm thinking I did something wrong. Should A be numerical? I don't see how . Thanks

2. Oct 22, 2013

### brmath

Probably a statistician would know this, but I don't. What is P(x)?

3. Oct 22, 2013

### Jbreezy

Probability. Idk. If they say find the probability that x <1 then you express it as P( x<1)

4. Oct 22, 2013

### Staff: Mentor

Since you aren't given a value for c, you aren't going to get a value for A that is a specific number. I don't see anything wrong in your work.

5. Oct 22, 2013

### Jbreezy

OK. So report my answer for P( x<1 ) = ∫ce^-ct dt = -e^(-ct) between [0,1]
as 1/e^c

6. Oct 22, 2013

### Staff: Mentor

I'm not sure you're writing what you mean.

$$\int e^{-ct}dt = -\frac 1 c \int e^{-ct} (-c dt) = -\frac 1 c e^{-ct} + C$$

For the definite integral you get 1/c. Note that I haven't included A, but if you multiply throughout by A, you get A/c. For that to be equal to 1, then A = c, which is what you had.

7. Oct 22, 2013

### LCKurtz

That isn't correct. $P(X<1)=\int_0^1 ce^{-cx}~dx$ and you have left out the lower limit in your calculation.

Also, I think most texts define the exponential density function as $\lambda e^{-\lambda x}$ for $x > 0$ in the first place.

8. Oct 22, 2013

### Jbreezy

This was my course pack. What do you mean it isn't right ? What isn't right? my answer for my integral 1/e^c?

I have the lower limit [0,1] what are you talking about.

9. Oct 22, 2013

### Jbreezy

OH I see the lower limit missing. But what else isn't right?

10. Oct 22, 2013

### Jbreezy

$P(X<1)=\int_0^1 ce^{-cx}~dx = 1-1/e^c$ Is this correct?

And is
$P(.5 < X<1.5)=\int_0^1 ce^{-cx}~dx = 1/e^c(0.5) - 1/e^c(1.5)$ correct?
I don't get what your saying is wrong LCKurtz

Last edited: Oct 22, 2013
11. Oct 22, 2013

### Staff: Mentor

I think so. It's been a long time since I studied mathematical statistics, and I seem to recall that this is a Poisson distribution.
$P(.5 < X<1.5)=\int_{.5}^{1.5} ce^{-cx}~dx$

In general, $P(a < X< b)=\int_a^b ce^{-cx}~dx$, where a ≥ 0 and b ≥ a.

12. Oct 22, 2013

### LCKurtz

Yes

Put the correct limits on the integral and it will be.

13. Oct 22, 2013

### Jbreezy

Oh I'm sorry. I just copied your tex and forgot to change it. It is OK on my paper.
Thanks

14. Oct 22, 2013

### Ray Vickson

The value A = c is correct. Also:
$$P(X \leq v) = 1 - e^{-cv}, \: \: P(X > v) = e^{-cv}.$$
These are used so often that if you plan to do more Probability in the future you should commit them to memory, along with
$$E X = \frac{1}{c}, \; \; \text{Var} X = \frac{1}{c^2}.$$

BTW: the standard way of writing is to use $e^{-cv}$ rather than $1/e^{cv}$. They are, of course, eqiuivalent, but the first way is used all the time by everybody, at least in works on probability.

Last edited: Oct 22, 2013
15. Oct 22, 2013

### Jbreezy

OH Ok cool thanks for the info ray.