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Math Help - Prove that f is the density function of a random variable Z?

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    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)
    Last edited by dkmathguy; April 12th 2010 at 06:20 PM.
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    Quote Originally Posted by dkmathguy View Post
    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 \int_0^{+\infty} x f(x) \, dx.

    3. Calculate \int_0^{+\infty} x^2 f(x) \, dx. Then 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.
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    1). f(x) takes only positive values and \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.
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    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?
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    Quote Originally Posted by dkmathguy View Post
    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)? \Pr(3 \leq X \leq 5) = \int_3^5 f(x) \, dx. Alternatively, use your answer to Q4: \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.
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    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.
    Last edited by dkmathguy; April 12th 2010 at 11:17 PM.
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    Quote Originally Posted by dkmathguy View Post
    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.

    F(x) = \int_0^x 2 e^{-2u} \, du for x \geq 0.
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    Quote Originally Posted by mr fantastic View Post
    Look, you have to go back and review this stuff. By defnition:

    F(x) = 0 for x < 0.

    F(x) = \int_0^x 2 e^{-2u} \, du for x \geq 0.
    So is it
    F(x) = 0 for x < 0
    F(x) = (e^(-2x))(e^(2x)-1) for x >= 0
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  9. #9
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    Quote Originally Posted by dkmathguy View Post
    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?
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    Quote Originally Posted by mr fantastic View Post
    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...
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