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Math Help - cdf question

  1. #1
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    cdf question

    I'm having a lot of trouble figuring out how to go about answering this question...

    I think part b) is just, Fx = integral of fx(x), and you'll get an expression for Fx...

    part a) would be something like..
    Fm = P(m <= x)
    = 1 - P(m > x)
    = 1 - Fx
    but I'm not sure how Fx and Fm relate together..

    As you can see, I'm pretty confused, so any help in starting off and understanding would be a great help!
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    Last edited by anonder; April 5th 2009 at 04:11 AM.
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  2. #2
    Moo
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    Hello,

    part a) would be something like..
    Fm = P(m <= x)
    = 1 - P(m > x)
    = 1 - Fx
    No !

    It is said that if M>x, then \forall i, X_i>x
    So :
    \mathbb{P}(M>x)=\mathbb{P}(X_1>x ,X_2>x,\dots,X_n>x)
    Since the Xi's are independent, you can write this as a product :
    \mathbb{P}(M>x)=\mathbb{P}(X_1>x)\dots\mathbb{P}(X  _n>x)
    Since they follow the same distribution, this is :
    \mathbb{P}(M>x)=[\mathbb{P}(X_1>x)]^n
    But \mathbb{P}(X_1>x)=1-\mathbb{P}(X_1 \leq x)=1-F_X(x)

    Hence we get :
    F_M(x)=1-[1-F_X(x)]^n

    I think part b) is just, Fx = integral of fx(x), and you'll get an expression for Fx..
    True ! But it's a bit more complicated for the details (not much, don't worry)

    And be careful, it's sometimes confusing if the dummy variable of the integral has the same name as one of the boundaries.

    F_X(x)=\int_{-\infty}^x f_X(t) ~dt
    Note that the density function is only defined for t>0.
    So if x<0, the integral is 0.
    If x>0, the integral is \int_0^x f_X(t) ~dt
    Can you understand why ?

    (don't forget to substitute \lambda with the value you're given)
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  3. #3
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    Thanks soo much Moo.

    Part a) makes so much sense now that you've stepped it out for me.

    For part c), i think you just need to substitute Fx from b) into Fx from the expression from a) to give Ft.
    F_T = 1 - [1 + exp(-\lambda x)[\lambda x + 1]]^n
    Is that the right method?

    For part d) I'm a bit lost on what they mean by let Y = root(n)T..
    Is Fy just Ft from part c) with Y = root(n)T? How do you use it?

    Thanks in advanced! Just pretty confused at all this..
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  4. #4
    Moo
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    Sorry, I forgot your thread

    For c) :
    For part c), i think you just need to substitute Fx from b) into Fx from the expression from a) to give Ft.
    F_T = 1 - [1 + exp(-\lambda x)[\lambda x + 1]]^n
    Is that the right method?
    Right method ! (because it is said that it goes out as soon as one light fails, so basically, the life time will be the min between the different lights' life time)
    But wrong formula..
    I get F_X(x)=1-e^{-\lambda x}(\lambda x+1) (for x>0)
    It seems that you didn't get the 1... If you have a problem in spotting where your mistake is (or if you think you're correct), provide what you did, and I'll help you go through it. But I'm confident you'll find it
    You can "check" it with the fact that the pdf is continuous -> the cdf is continuous -> F_X(0) should be equal to 0 (according to the previous question)

    And F_T would then have a more friendly expression ^^



    For d) :
    F_Y(y)=\mathbb{P}(Y \leq y)=\mathbb{P}(\sqrt{n} T \leq y)=\mathbb{P}\left(T \leq \frac{y}{\sqrt{n}}\right)

    So what does it look like ? Can you relate it to F_T(t)=\mathbb{P}(T\leq t) ?



    For e) :
    I don't like it... If it has to be related to d), it means that you have to consider 50 as a large number (since you had the limit as n goes to infinity)...
    As an answer to this question, you can "see" that the limit of F_Y(y) looks like the cdf of an exponential distribution. But for the moment, I don't see how to get E(T)
    Last edited by Moo; April 5th 2009 at 04:35 AM.
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