# Find the mode of a probability distribution function

I am trying the find mode of a probability distribution function given by

$$g(x/\alpha,\beta,\sigma)=\frac{1}{\Gamma \left( \alpha \right)\beta^{\alpha}}exp\left\{{-\frac{x^2}{2\sigma^{2}}\frac{1}{\beta}}\right\}\frac{x^{2\alpha-1}}{2^{\alpha-1}\sigma^{2\alpha}}I_{{\rm I\!R}^{+}}(x)$$ where $\alpha, \beta, \sigma >0$.

The limit of this density function as $x \to \infty$ is $0$ and the limit of its hazard function as $x \to \infty$ is $\infty$. Also the limit of this density function and its hazard function as $x \to 0^{+}$ is given by

$$\lim_{x \to 0^{+}} g(x)=\lim_{x \to 0^{+}} h(x) = \begin{cases} 0 & \alpha > \dfrac{1}{2} \\ \dfrac{\sqrt{2}}{\sigma\sqrt{\pi \beta}} & \alpha= \dfrac{1}{2} \\ \infty & \alpha < \dfrac{1}{2} \end{cases}$$

The mode can be obtained by taking the derivative of $g(x)$ and setting it to zero. The obtained mode is an increasing function of $\alpha$, $\beta$ and $\sigma$. Thus, The mode given as $$k(x)=\sigma \sqrt{ \beta(2\alpha -1)}.$$

However, the definition of square root, the mode is only defined when $\alpha \geq \dfrac{1}{2}$. However, my $\alpha$ parameter is defined on $(0, \infty)$. There's something I am missing out!!

• The mode is the peak of the distribution. Since you have constraints on parameters, you should be careful with simply taking a derivative and equaling it to zero. Use Kuhn-Tucker conditions. – Aksakal Dec 25 '14 at 23:30
• What you are missing is actually staring you in the face in what you have written in your question, namely, $\lim_{x \to 0^{+}} g(x)= \infty$ when $\alpha < \frac 12$. So, since $g(x)$ is finite on $(0,\infty)$, it cannot have a mode (that is, maximum value) in $(0,\infty)$ – Dilip Sarwate Dec 25 '14 at 23:49
• The formula for $k$ is incorrect when $2\alpha-1\lt 0$. Suppose (for simplicity) that $\beta=\sigma=1$. The mode can occur only in the limit as $x\to 0$ or when $-x + (2\alpha-1)/x = 0$ and $x\gt 0$. The latter has no solution when $2\alpha-1\lt 0$. – whuber Dec 26 '14 at 13:53
• It is actually not correct. This is what we are trying to solve. – ARAT Dec 26 '14 at 14:06

Consider that there are shapes of pdf that have a mode, but at which the derivative of the pdf is not zero (the Laplace being an obvious example).

There are also cases where there's no mode in the domain of the variable (examples below).

That is, we can't say as a general statement "the mode can be obtained by taking the derivative of $g(x)$ and setting it to zero".

You must give up trying to look at the value where the derivative is zero when it's clear that the reasoning underlying its use (that the derivative is zero at the mode) fails. Here's a simple example where it works for part of the domain of a parameter, but not everywhere:

(Here I've include $x=0$ in the domain for the exponential case; if it was excluded, strictly there's no mode in the domain for that case (the supremum would be $1$, which is the value of $\frac{1}{\mu} \exp(-\frac{x}{\mu})$ at $0$, but the density would not be defined there). For the case where the shape parameters $<1$, you couldn't include $x=0$, and there would be no mode.)

So there's no use looking for where the derivative is zero when it isn't zero at the mode - indeed it may not be zero anywhere.

Further, for some densities, even when the derivative is 0, it doesn't imply there's a mode there. Consider this density (a beta density):

There's not a local mode where the derivative is zero; it's an antimode. You can also have a density with a horizontal point of inflexion which will be neither a mode nor an antimode:

As a result, it's not sufficient to simply calculate a formula at which the derivative is zero; even if you can calculate such values, that may not tell you where the modes are.

You must work out when that calculation corresponds to modes of the density; where that fails, quite often the location of the mode is obvious (if it exists at all).

A first exercise would be to draw the density at a few values of $\alpha$ and see how it behaves. Then you should be able to bring some reasoning to bear on the problem, which will tell you where the mode is when $\alpha <\frac{1}{2}$.

It's possible for the function to be everywhere decreasing in the domain; if it's on an open interval there may be no mode in that interval. For example, in the case of the gamma with $\alpha<1$, it's not uncommon to say "there's a mode at 0" even though the limit isn't in the interval - it's strictly incorrect to say there's a mode in that case (but usually one can understand the actual intent if someone says there's a mode at 0 even though the function is unbounded in the limit).

Here's an example of what your density looks like at $\beta=\sigma=1$ for three different values of $\alpha$ near $0.5$:

It does clearly suggest the behavior for $\alpha<\frac{1}{2}$ may be monotonically decreasing. The obvious thing to do is to try to check whether it's the case that $g(x+\epsilon)<g(x)$ with $\alpha<\frac{1}{2}$ when $\epsilon>0$ (e.g. is the derivative always negative when $\alpha<\frac{1}{2}$?)

If it's the case that it is monotonic decreasing (you should carry out such a check for yourself), it doesn't strictly have a mode; best to simply describe the behavior near 0 for $\alpha<\frac{1}{2}$.

• Thank you Glen_b, I think I understand pretty much of what you wrote. That was what concerns me the most about finding the mode of a density, taking derivative of function and setting it to zero. Because there're many articles who do that, e.g. tandfonline.com/doi/abs/10.1080/03610926.2011.599002 So, I need to be extra careful when $\alpha< \dfrac{1}{2}$. But what about when $\alpha \geq \dfrac{1}{2}$? I have $k(x)=\sigma \sqrt{ \beta(2\alpha -1)}$ and is not it considered for mode? As Dilip Sarwate commented, is it just a maximum value? – ARAT Dec 26 '14 at 9:51
• I have added the plots of this density for fixed $\alpha \geq \dfrac{1}{2}$ and $\sigma$ with different values of $\beta$. What I want to say when I draw the graph, this density seems to have a mode. – ARAT Dec 26 '14 at 9:53
• Those modes you see are definitely modes, and they're where you think they are by the formula. You should consider what it takes to prove you have a local mode (this is covered in basic calculus) and what else is needed to show that it's the only one (i.e. that there are no other modes). [It's not especially complex, just a little more is involved than simply computing where the derivative is 0. You're nearly half-way done in what's needed.] – Glen_b Dec 26 '14 at 10:44
• You'll notice if you read the paper you link to carefully that they show that the density is unimodal (among other things) -- that is, they do some of things I was discussing in my earlier comment. – Glen_b Dec 26 '14 at 10:56
• @MuratArat If $g(x)$ is differentiable on $(0,\infty)$, $g(0) = 0$ and $g'(x)$ has a unique zero $m$ in $(0,\infty)$, then $m$ is necessarily the mode of $g(x)$ since all densities necessarily approach $0$ as $x \to \infty$. – Dilip Sarwate Dec 26 '14 at 14:26