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Good Morning.

I have recently been faced with modeling a quantity that is best modeled via a Gamma distribution. I have noticed that, in the characterization of the distribution via the parameters $k$ and $θ$, the mean of the distribution is $kθ$, whereas the "top" of the distribution, representing the point that is sampled with the highest probability, is found at point $θ( k - 1)$.

This contradicts my intuition. If the mean represents the expected value of the distribution, then why is it not placed at the "top" of the distribution, much like in the normal distribution?

Thanks,

Jason

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The "top" of the distribution represents the one most likely outcome, and is called the mode of the distribution. This is different from the expected value, which is what the average outcome would be after a large number of trials. The gamma distribution is asymmetrical, and such distributions usually have a mode that differs from the expected value.

Let's take a simpler example. Suppose I roll a die and give you a dollar if it comes up 1, 2, 3, 4, or 5, but if it comes up 6 you have to give me \$100. The most likely outcome is that you get a dollar. But if we keep playing this game, in the long run you will tend to lose money. In terms of the probability distribution of your profit per game, its mode is \$1, but its expected value is −\$15.83.

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    That's the essence of the work that I (and my colleagues) have been carrying out. More or less, we've ported a Prolog - based activity recognition system to a probabilistic logic programming framework which, unlike a Prolog-based system, acknowledges probabilities in its input. The paper describing this system is under review for publication by the TPLP journal but I'm able to refer people to the technical report version here: http://arxiv.org/abs/1204.1851. I'm looking into ways to refine the probability generation described in section 8 of that paper because it is done rather crudely.2012-05-10
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How about this: $ X = \begin{cases} 1 & \text{with probability } 1/5 \\ 2 & \text{with probability } 2/5 \\ 3 & \text{with probability } 1/5 \\ 4 & \text{with probability } 1/5 \end{cases} $

Then the "top"---the most probable outcome, generally called the "mode"---is 2, and the mean or expected value is $\mathbb{E}(X)=2.4$