1. ## method of moments

f(y| $\theta$)= ${e}^{[-(-(y-\theta))]}$, y≥ $\theta$ where theta is an unknown positive constant.

Find a moment estimator for theta and adjust theta hat to find an unbiased estimator for theta.

Solving for E(Y) I get $y{e}^{y-x}-{e}^{y-x}+{e}^{-x}$ which is my sample average (Y-bar).

Am I on the right track?

2. I cannot understand your exponent.
It looks like absolute values, but I doubt you want that.
I bet you mean $e^{-(y-\theta)}$ when $y\ge \theta$.

3. That's right. Those are supposed to be braces

4. BUT you have 2 negative signs, which means your exponent blows up.
Hence you have infinite area.

5. yes the integration used to obtain E(Y) goes from $/theta$ to infinity. Oops I was going from 0 to y. This is tricky. I have two other problems related to this one. An explanation of this problem will hopefully help with the others.

6. PLEASE type the correct density and the bounds on the variable.
I don't know why you have 2 negative signs.

7. ok my hw has a typo. Just got confirmation its e^(-(y-theta)) for the density function.

8. now I have ye^(-y+x) + e^(-theta + x) as my E(Y)

not exactly sure what the next step would be.

9. setting the above equal to Ybar and solving for theta I get theta = -ln(Ybar-ye^(-y+x)) -x

10. Originally Posted by Snarf?
ok my hw has a typo. Just got confirmation its e^(-(y-theta)) for the density function.
I stated that several hours ago.

11. This is getting weird.
What's x now???
ye^(-y+x) + e^(-theta + x) does not make sense.
I BET you'e "integrating" x times f(y).
THERE is only ONE variable.
Either use x or y, it doesn't matter, these are dummy variables.
JUST integrate yf(y) from theta to infinity.
When you integrate y and plug in bounds, the y is no longer there.
And this x is purely fictional.
you need to review calculus.
You should have ONLY a function of theta as the expectation of Y and you set that to the sample mean.

12. Somewhere along the way during integration by parts I substituted x for theta and forgot to change it. I do that type of thing a lot. Anywho, now I have ye^(-y+theta) + 1 for my E(Y)

13. Originally Posted by Snarf?
Somewhere along the way during integration by parts I substituted x for theta and forgot to change it. I do that type of thing a lot. Anywho, now I have ye^(-y+theta) + 1 for my E(Y)
I assume you're using an integral calculator.
No wonder I ban them in my classroom.
ONCE again PLUG in your bounds.

14. the bounds I used are from theta to infinity. Is this correct?

integration by parts:

ye^(-y+theta) - integral from theta to infinity of e^(-y+theta) dy

= ye^(-y+theta) -(e^(-y+theta)| (from theta to infinity)

= ye^-(y+theta)-(0-e^(-theta + theta)

15. separate the exponent e^theta and e^(-y) and PULL that outside the integral
then integrate ye^(-y) via parts and plug in theta and let y go to infinity for your bounds.
WHEN you're done there can't be any more y's, only theta's.
Then set that equal to the sample mean.
AND
e^(-theta + theta) =e^0=1

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