I am toying with mixture models, especially in a bayesian context and the Gamma (or the inverse Gamma) distribution appears quite often. For example, inverse Gamma is used as a conjugate prior for the variance of a gaussian mixture. The shape parameter seems to have a quite intuitive interpretation. However, I am more confused with the rate parameter. Is there any clear interpretation for the rate parameter (or inverse scale) of a Gamma distribution ?
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$\begingroup$ Gamma distributions can be parameterised in terms of mean and variance, if it helps. $\endgroup$– Xi'anCommented Jan 20, 2019 at 12:10
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Although there are certainly other ideas, here is my quite long intuition about the Gamma Distribution and its parameters:
When we sum $\alpha$ (integer) independent exponential RVs, with rate parameter $\beta$, we obtain a Gamma RV with shape $\alpha$ and rate $\beta$. Of course, Gamma distribution is also generalized to the cases where $\alpha$ is not an integer, but I believe this won't be significant barrier for us to build up intuitions. So, as we increase the shape, $\alpha$, as the number of exponentials being summed increases, the PDF moves towards right because the expected value increases, as well as the probability of getting larger numbers since the expectation of each exponentials stays the same, i.e. rate, $\beta$, is fixed.
The rate parameter belongs to the hypothetical exponential RVs being summed. As it increases, the PDF becomes steeper, i.e. $\beta e^{-\beta x}$; as it decreases, the PDF becomes to have heavier tails, increasing the possibility that the exp. random variable having higher values. These heavier tails also increase the variance of the Gamma distribution, while pushing Gamma PDF to the right. So, the arms of the Gamma PDF gets larger and larger as the rate parameter cools down.
The exponential RV (so Gamma) has an intimate relationship with Poisson RV. One very common use case, and a very intuitive one, of Poisson RV is modelling the number of events occurring in a particular time interval, e.g. $X:$ number of customers visiting a store in a given time interval (e.g. an hour) is a Poisson RV with parameter $\beta$. here, this parameter is also called rate, and it is also the expectation of Poisson RV, which is equivalent to saying that we on average expect $\beta$ number of customers in an hour. When we make this definition, and record each customer's arrival on a time axis (with independent customers assumption), the inter-arrival times suddenly become exponential RVs. So, the time between any two consecutive customer is an exponential RV with rate, $\beta$. Exponential mean is $\frac{1}{\beta}$. It's like saying if we expect $\beta$ customers in an hour, then average time between their arrivals will be $\frac{1}{\beta}$ hour. What is the total time to needed to have $\alpha$ number of customers? It is the sum of $\alpha$ exponentials with rate $\beta$, i.e. Gamma RV.
So, $\beta$ can be thought of as event occur rate.
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$\begingroup$ How do you know this: the PDF moves towards right because the expected value increases ? $\endgroup$ Commented Jan 24, 2019 at 2:38
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$\begingroup$ Good point! Firstly, in that paragraph, $\beta$ is fixed. And, by moving right, I mean graphically seeing the PDF moving right, an illusion which is created by the movement of the mode of the PDF moving right. The mean is $\alpha/\beta$, and mode is $(\alpha-1)/\beta$ for $\alpha \geq 1$. As mean increases, the mode increases. As the explanation is about summing multiple exponentials, i.e. $\alpha\geq 1$, the relation holds. $\endgroup$– gunesCommented Jan 24, 2019 at 10:14