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    How do I construct the inverse of 2x-5t?
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    What have you tried so far?
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    What I've tried is to construct the inverse of the exponential to give a new pdf, but I can't get the inverse...
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    First find the c.d.f. F for f(x). That is, find p(X <= x).

    Then if Y is unif(0,1), F^{-1}(Y) has the same distribution as X.
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    (Original post by DFranklin)
    First find the c.d.f. F for f(x). That is, find p(X <= x).

    Then if Y is unif(0,1), F^{-1}(Y) has the same distribution as X.
    So I'm to integrate it first... got it.
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    (Original post by DFranklin)
    First find the c.d.f. F for f(x). That is, find p(X <= x).

    Then if Y is unif(0,1), F^{-1}(Y) has the same distribution as X.
    I've got the integral as F(x)=-exp(-x^3), but I can't find the inverse of it. Help?
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    Well, what's the inverse of exp?
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    (Original post by DFranklin)
    Well, what's the inverse of exp?
    This is where I get stuck:

    U = -exp(-x^3)
    -U =exp(-x^3)
    ln(-U) = -x^3
    -ln(-U) = x^3
    cube root of (-ln(-U)) = x

    But this looks nothing like other examples I've covered, in which x is usually ln(1-U) or something simple
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    Check your arbitrary constant (from when you integrated). What is P(X<=infinty) going to be?
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    (Original post by DFranklin)
    Check your arbitrary constant (from when you integrated). What is P(X<=infinty) going to be?
    Sorry, I don't understand :o: When I put the limits 0 and infinite in, the integral works out as 1. :confused:
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    Who said anything about limits (definite integral limits, at any rate)?

    By definition, F(x) = P(X <= x).

    So F(\infty) had better equal 1.

    Since -exp(-x^3) doesn't tend to 1 as x \to \infty, you have a (easily remedied) problem...
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    (Original post by DFranklin)
    Who said anything about limits (definite integral limits, at any rate)?

    By definition, F(x) = P(X <= x).

    So F(\infty) had better equal 1.

    Since -exp(-x^3) doesn't tend to 1 as x \to \infty, you have a (easily remedied) problem...
    Oh! Is it to change it to 1 -exp(-x^3)? :o:
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    Yes. After that, your method should be OK, I think.
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    (Original post by DFranklin)
    Yes. After that, your method should be OK, I think.
    Oh. Huge thanks for helping me through this question step-by-step. :yes:
 
 
 
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