# Thread: Prove that f is the density function of a random variable Z?

1. ## Prove that f is the density function of a random variable Z?

f(x) =
0 ............... if x <= 0
2e^(-2x) .... if x > 0

1) Prove that f is the density function of a random variable Z.
2) Compute E(Z).
3) Compute V(Z) and standard deviation(Z)
4) Find the distribution function of Z.
5) Compute P(3 <= X <= 5)

2. Originally Posted by dkmathguy
f(x) =
0 ............... if x <= 0
2e^(-2x) .... if x > 0

1) Prove that f is the density function of a random variable Z.
2) Compute E(Z).
3) Compute V(Z) and standard deviation(Z)
4) Find the distribution function of Z.
1. Check that the two criteria for f(x) to be a pdf are satisfied.

2. Calculate $\displaystyle \int_0^{+\infty} x f(x) \, dx$.

3. Calculate $\displaystyle \int_0^{+\infty} x^2 f(x) \, dx$. Then $\displaystyle Var(X) = E(X^2) - [E(X)]^2$.

4. Proceed in a similar way to finding the distribution function in your other thread.

If you need more help, please show all your work and say where you get stuck.

3. 1). f(x) takes only positive values and $\displaystyle \int_{-\infty}^{\infty}f(x)dx=2\int_0^{\infty} e^{-2x}dx=-0.5e^{-2x}|_0^{\infty}=1$. So it is a density function.

4. 1) f(x) takes only positive values and . So it is a density function.

2) Compute E(X).
Integral from 0 to infinity of (x)(2e^(-2x)) dx.
I got 1/2, is that correct?

3) Compute V(X) and standard deviation(X)
Integral from 0 to infinity of (x^2)(2e^(-2x)) dx.
I got 1/2, is that correct?
1/2 - (1/2)^2 = 2/4 - 1/4 = 1/4?

standard deviation(X) = 1/2?

4) Find the distribution function of X
Integral of (2e^(-2x))?
-e^(-2x)?

5) Compute P(3 <= X <= 5)
mean = 1/2 ............ standard deviation(X) = 1/2
(3-0.5)/0.5 = 5
(5-0.5)/0.5 = 9
P(x > 5) - P(x > 9).... ummm, huh?

5. Originally Posted by dkmathguy
1) f(x) takes only positive values and . So it is a density function.

2) Compute E(X).
Integral from 0 to infinity of (x)(2e^(-2x)) dx.
I got 1/2, is that correct?

3) Compute V(X) and standard deviation(X)
Integral from 0 to infinity of (x^2)(2e^(-2x)) dx.
I got 1/2, is that correct?
1/2 - (1/2)^2 = 2/4 - 1/4 = 1/4?

standard deviation(X) = 1/2?

4) Find the distribution function of X
Integral of (2e^(-2x))?
-e^(-2x)?

5) Compute P(3 <= X <= 5)
mean = 1/2 ............ standard deviation(X) = 1/2
(3-0.5)/0.5 = 5
(5-0.5)/0.5 = 9
P(x > 5) - P(x > 9).... ummm, huh?
1, 2 and 3 are OK.

4 is done like the one I did for you in the other thread. Follow that thread and be specific with what you don't ubnderstand.

5. What on Earth are you doing (actually I know what you're doing but I have no idea why you're doing it)? $\displaystyle \Pr(3 \leq X \leq 5) = \int_3^5 f(x) \, dx$. Alternatively, use your answer to Q4: $\displaystyle \Pr(3 \leq X \leq 5) = F(5) - F(3)$.

You need to seriously review your class notes and textbook because this is all basic stuff that follows from definitions and formulae.

6. 4) Find the distribution function of X
Integral of (2e^(-2x))?
F(x) = 0 ..........for x <= 0
F(x) = -e^(-2x) for x > 0

I am confuse on this one because it doesn't have a c to solve for, x > 0 instead of 0 < x < ?, and it has a e^x in it.

5) Oops, I was trying to found z = (X - μ) / σ.
Compute P(3 <= X <= 5)
F(5) = -e^(-10)
F(3) = -e^(-6)
[-e^(-10)] - [-e^(-6)]
e^(-6) - e^(-10) = .0024

Are these correct? Thanks again Mr. Fantastic, no wonder you are Mr. Fantastic.

Sorry about not knowing my material, I decided to major in statistic(economy might get worst in 2012 so I didn't want to be stuck with a useless degree) late in my junior year, so I am currently taking 3 math classes to catch up, plus, there is this hottie in my statistic class and that just makes it harder to concentrate on the professor lecture.

7. Originally Posted by dkmathguy
4) Find the distribution function of X
Integral of (2e^(-2x))?
F(x) = 0 ..........for x <= 0
F(x) = -e^(-2x) for x > 0

I am confuse on this one because it doesn't have a c to solve for, x > 0 instead of 0 < x < ?, and it has a e^x in it.

[snip]
Look, you have to go back and review this stuff. By defnition:

F(x) = 0 for x < 0.

$\displaystyle F(x) = \int_0^x 2 e^{-2u} \, du$ for $\displaystyle x \geq 0$.

8. Originally Posted by mr fantastic
Look, you have to go back and review this stuff. By defnition:

F(x) = 0 for x < 0.

$\displaystyle F(x) = \int_0^x 2 e^{-2u} \, du$ for $\displaystyle x \geq 0$.
So is it
F(x) = 0 for x < 0
F(x) = (e^(-2x))(e^(2x)-1) for x >= 0

9. Originally Posted by dkmathguy
So is it
F(x) = 0 for x < 0
F(x) = (e^(-2x))(e^(2x)-1) for x >= 0
Yes. But why have you written it in such a complicated way?

10. Originally Posted by mr fantastic
Yes. But why have you written it in such a complicated way?
Oh, my calculator pop out that answer.
F(x) = 0 for x < 0
F(x) = 1 - e^(-2x) for x >= 0

Thanks again, I think I am going to stalk around these forum to learn some more stuff. Usually, when I have a problem and have the answer beforehand, I can figure out how to solve the problem. But your method of not giving me answer and instead give me hints, really help me understand better. I will most certainly past down my knowledge from you in the future...