Calculus - Minimizing functions

[cutesteph]

Homework Statement

Maximize the function for x. exp(-y+x-1) for y >= x-1 and y=x+e where e is distributed exp(L)

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.

 

[LCKurtz]

Homework Statement

Maximize the function for x. exp(-y+x-1) for y >= x-1 and y=x+e where e is distributed exp(L)

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.

What does "maximize the function for x" mean? And what is L? And what does "e is distributed exp(L) mean?

 

[cutesteph]

What does "maximize the function for x" mean? And what is L? And what does "e is distributed exp(L) mean?

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).

 

[LCKurtz]

Basically I am finding the maximum likelihood estimator x which is equivalent to maximizing the function exp(-y+x-1) for y >= x-1 .

e is representing error and distributed exponentially with parameter L means that it follows a distribution L*exp(-L*x).

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.

 

[Ray Vickson]

Homework Statement

Maximize the function for x. exp(-y+x-1) for y >= x-1 and y=x+e where e is distributed exp(L)

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.

Are you describing two separate problems? I read it as:
Problem (1) \max_x \exp(-y+x-1),\\<br /> \text{subject to } x-1 \leq y
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 $x = +\infty$.