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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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2 Answers 2

up vote 3 down vote accepted

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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Thanks, I got it. Here's another relevant question: I have been using the distribution's CDF (link to exact characterization: en.wikipedia.org/wiki/Gamma_Distribution) to make sure that my parameters k and θ are such that 98% of the distribution's mass is between 0 and 1. Moreover, I want to vary either the mean or the mode so that the distribution tends to produce values closer to 0 or to 1. Given this short discussion, would you prefer to constrain the mode so that it is close to 0 (or 1) or the mean to be close to 0 or 1? Mathematically, θ*(κ-1) =0.1 or θ*k = 0.1; –  Jason May 10 '12 at 9:12
    
Given the shape of the gamma distribution, I don't see how you can get either of them anywhere close to 1 while simultaneously keeping 98% of the mass between 0 and 1... unless $k$ is really high, in which case there won't be much of a difference between $\theta k$ and $\theta (k-1)$. Can you elaborate on what you are really trying to achieve by varying the parameters of the distribution? –  Rahul May 10 '12 at 9:23
    
Absolutely. I'd like to model noise in the form of probabilities. I'd like to represent different levels of noise, some more severe (probabilities close to 0), some less (probabilities close to 1). What I've been doing is constrain the CDF as pre-mentioned (so that I don't compromise the distribution's shape) and set the mode to 0.1, 0.2, 0.3 and 0.4. These choices will give me low probabilities; when it's time to obtain higher probabilities, I "mirror" the distribution at 0.5 by sampling the 1-complement of these values. My question is:do I set the mode or mean to those hardcoded values? –  Jason May 10 '12 at 9:37
    
If you will notice, the constraining of the CDF as well as the hardcoding of either the mean or the mode to 0.1, 0.2 and so on give me a system of 2 equations, which I can then solve to obtain both k and θ. –  Jason May 10 '12 at 9:39
    
Actually, the more I think about it, the more my question seems self-answerable from the semantics of the mean and the mode themselves: Since I will need to sample a bunch of random numbers from every single PDF, it is the mean, and not the mode, that I want to revolve around from. –  Jason May 10 '12 at 9:46

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 $\mathbb{E}(X)=2.4$

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