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For standard normal Z, the change in probability density associated with equally sized changes in z is obviously greater for values of z that are further away from the mean/mode.

For example, if z=0, then $\phi(z)- \phi(z+0.1)\approx 0.002$. However, if z=2, then $\phi(z)-\phi(z+0.1) \approx 0.01$. Thus the density change associated with increasing z by 0.01 is 5 times as large in the latter case.

I have two questions related to this.

  1. Does this property (or a similar property) have a name? (That is, the property where for rv X with PDF f and mean value $\bar{x}$: $\frac{\partial f(x)}{\partial x}<\frac{\partial f(x^o)}{x^o}$ when $\lvert x^o- \bar{x} \rvert >\lvert x- \bar{x} \rvert$.) Does this general idea - that is the extent to which the PDF's derivative changes over the distribution's support - have a name?

  2. Are probability distributions with this property (or a similar property) systematically classified? Can they be identified in some straightforward way? Obviously, all normal distributions have it, but do all unimodal continuous finite distributions?

I am asking because I am working with a finding that is dependent on the above property and I'd like to figure out the best way to succinctly talk about it, and also how to accurately think and report about its generality (or lack thereof).

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    $\begingroup$ The property as stated does not hold for the standard normal for all z (try $z=4$). Furthermore, if the domain of $f$ is not bounded, wouldn't this imply that $f$ is negative for some $x$? Please clarify if there is some mistake in stating the property you are thinking of. $\endgroup$ – Juho Kokkala Aug 5 '16 at 5:43
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    $\begingroup$ Assuming "elasticity" is a reference to the concept in economics, note that that is the ratio of two percentage changes, not just eg "change in demand per change in price". So I'm not sure in what sense "elasticity" is being used here. $\endgroup$ – Silverfish Aug 5 '16 at 10:16
  • $\begingroup$ If you are talking about changes in the slope of the pdf, more relevant ideas might be related to the second derivative and convexity and concavity $\endgroup$ – Silverfish Aug 5 '16 at 10:17
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Note (proceeding somewhat loosely and assuming the necessary derivatives exist and so on for it all to work) that $f(x+\delta)\approx f(x) + \delta f'(x)$ (e.g. consider a first order Taylor expansion) and so the rate of change $\frac{f(x+\delta)-f(x)}{x+\delta-x}\approx f'(x)$. Indeed in the limit as $\delta$ becomes small this will become the result.

So you're effectively saying "why is $|\phi'(x)|$ at $x=0$ smaller than elsewhere?".

The answer is that the function $\phi(x)$ is flat when you're at the top of the hill (any mode of a continuously differentiable density):

standard normal density

The normal has only one mode. The derivative is $0$ there, and non-zero everywhere else. Let's look at a plot of $|\phi'(x)|$:

absolute derivative of standard normal density

Clearly it's bigger everywhere than at the mode (= mean = 0).

This function (the absolute value of the derivative of the density) doesn't have any particular name I am aware of.

The absolute derivative of any other normal will follow the same pattern about its mode.

Obviously, all normal distributions have it, but do all unimodal continuous finite distributions?

No. First we have to get away from saying mean/mode. In general the mean and mode aren't in the same place. The mean may well be situated in a place where the derivative is large. So lets focus on the mode.

  • It's perfectly possible to have a unimodal density where the derivative is 0 not at a mode.

    ![density with "flat" section away from the mode

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  • $\begingroup$ This is helpful in showing that there are unimodal finite distributions where this property doesn't hold and in indicating that there is no standard approach to describing it. I think perhaps I should have picked a different starting illustrative example though. I am no so much asking about why the absolute derivative of a Normal PDF is smallest at the mode but more about the types of distributions in which the absolute derivative of the PDF will monotonically increase in absolute distance from the mode. Does that make sense? Any additional thoughts? $\endgroup$ – BGTP33 Aug 5 '16 at 7:48
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    $\begingroup$ But the normal is not an example of it monotonically increasing with distance from the mode. See my 2nd plot which shows that it increases until $\pm 1$ sd either side of the mode, then decreases, asymptotically approaching 0. If you want to ask about cases unlike normal where it might continue to increase you can do that, but there will be an infinity of such distributions. The beta(2,2) would be a canonical example of one $\endgroup$ – Glen_b Aug 5 '16 at 7:51
  • $\begingroup$ That's helpful. I missed the point of your second graph when I first read response - got it now though. Thanks for your time. $\endgroup$ – BGTP33 Aug 5 '16 at 16:48

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