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Calculus - Minimizing functions

  1. May 20, 2014 #1
    1. The problem statement, all variables and given/known data
    Maximize the function for x. exp(-y+x-1) for y >= x-1 and y=x+e where e is distributed exp(L)


    3. The attempt at a solution

    d/dx(ln(exp(-y+x-1)) = 0 => d/dx(-y+x-1) = 0 but if I take the derivative of this x goes away.
     
  2. jcsd
  3. May 20, 2014 #2

    LCKurtz

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    What does "maximize the function for x" mean? And what is L? And what does "e is distributed exp(L) mean?
     
  4. May 20, 2014 #3
    Basically I am finding the maximum likelihood estimator x which is equivalent to maximizing the function exp(-y+x-1) where y >= x-1 .

    e is representing error and distributed exponentially with parameter L means that it follows a distribution L*exp(-L*x).
     
  5. May 20, 2014 #4

    LCKurtz

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    Well, we aren't mind readers here. How would we know that "e" in your problem isn't the base of natural logarithms? You should include relevant details in your statement of the problem. I will let others respond to your question, now that I know what the subject area is.
     
  6. May 20, 2014 #5

    Ray Vickson

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    Are you describing two separate problems? I read it as:
    Problem (1) [tex] \max_x \exp(-y+x-1),\\
    \text{subject to } x-1 \leq y [/tex]
    and
    Problem (2) ##\max_x x+e, \: e \sim \text{expl}(L)##

    If so, you have approached problem (1) incorrectly, since you cannot just set the derivative to zero in a constrained problem.

    As stated, problem (2) makes no sense, for at least two reasons: (i) the thing you are maximizing is a random variable, not a real-valued function; and (ii) if there is no constraint on ##x## your "maximum" will be at ##x = +\infty##.
     
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