Turn on thread page Beta
    • Thread Starter
    Offline

    0
    ReputationRep:
    How do I construct the inverse of 2x-5t?
    Offline

    18
    ReputationRep:
    What have you tried so far?
    • Thread Starter
    Offline

    0
    ReputationRep:
    What I've tried is to construct the inverse of the exponential to give a new pdf, but I can't get the inverse...
    Offline

    18
    ReputationRep:
    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.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (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.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (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?
    Offline

    18
    ReputationRep:
    Well, what's the inverse of exp?
    • Thread Starter
    Offline

    0
    ReputationRep:
    (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
    Offline

    18
    ReputationRep:
    Check your arbitrary constant (from when you integrated). What is P(X<=infinty) going to be?
    • Thread Starter
    Offline

    0
    ReputationRep:
    (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:
    Offline

    18
    ReputationRep:
    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...
    • Thread Starter
    Offline

    0
    ReputationRep:
    (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:
    Offline

    18
    ReputationRep:
    Yes. After that, your method should be OK, I think.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (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:
 
 
 

1,112

students online now

800,000+

Exam discussions

Find your exam discussion here

Poll
Should predicted grades be removed from the uni application process
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.