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Given this probability density function $f(y|\theta)=(\frac{1}{\theta^2})ye^{\frac{-y}{\theta}}, y>0$ and $0$ otherwise. I found that my maximum likelihood estimator, by taking the derivative of the log of the likelihood function, would be $\Theta=\frac{1}{2}\bar{Y}$ but I also found that by taking the second derivative that my max will occur at $\Theta<\bar{Y}$. What's the significance of this? or what does it mean?

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I don't see any contradiction, nor anything that would require explanation.

There's a single turning point in the log-likelihood, which you identified. The second derivative at that turning point is negative.

There's also a point of inflexion at $\theta=\bar{y}$.

What's the difficulty?

Did you try drawing the log-likelihood for some small sample? It might help you. Here's a sample to try if you want one:

1.76 3.10 1.32 1.93 2.71 3.34 0.65 0.24 0.25 3.19

Edit: Here's the benefit of drawing a picture -- it all becomes much clearer:

Drawing of log likelihood for given sample, showing maximum at mean/2 and point of inflexion at mean

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  • $\begingroup$ Well taking the second derivative tells you whether the function reaches its maximum at the critical point if it's less than 0. Since we are working with MLE's, it just appeared to me a little weird I was able to deduce that the estimator reaches it's maximum when it is strictly less than the $\bar{Y}$, I find it counterintuitive since $\bar{Y}$ is usually the most sufficient estimator for most functions $\endgroup$
    – Jaider
    Commented Nov 13, 2016 at 13:47
  • $\begingroup$ But looking at the function and the MLE of $1/2\bar{Y}$, it kind of makes sense now $\endgroup$
    – Jaider
    Commented Nov 13, 2016 at 13:50
  • $\begingroup$ Ah, scalar multiplication of $\bar{Y}$ makes it unbiased for this function $\endgroup$
    – Jaider
    Commented Nov 13, 2016 at 14:24

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