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I've been trying to wrap my head around different types of pdf.

Do pdfs show relative probabilities?

Take this example of the exponential:

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When lamba=1.5 p(0)=1.4

Obviously the probability of something cannot be 1.4

Is this a relative probabilty? We can estimate that p(1)=0.3

is the event (x=0) 1.4/0.3 = 4.67 times more likely to occur than the event (x=1) ?

I suppose I don't really know what a pdf really IS

Is the reason you get values higher than 1.0 on the y axis just a consequence of scaling the pdf so the total area is 1?

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    $\begingroup$ A density function is not a probability. $\endgroup$ Commented Jun 1, 2016 at 0:21
  • $\begingroup$ That's what I'm asking - what is it? $\endgroup$
    – RNs_Ghost
    Commented Jun 1, 2016 at 0:22
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    $\begingroup$ If $\Delta x$ is small, then $(\Delta x)f(x)$ approximates well the probability that $X$ is between $x$ and $x+\Delta x$. So $(\Delta x)f(x)$ is a probability. $\endgroup$ Commented Jun 1, 2016 at 0:25
  • $\begingroup$ thanks - but I'm afraid that really doesn't help me understand what a pdf is. Is f(x) the y axis of the pdf? $\endgroup$
    – RNs_Ghost
    Commented Jun 1, 2016 at 0:29
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    $\begingroup$ The subtle issue is that the value of the pdf at one point does not mean anything. If you change the value of a pdf at any one point, or finitely many points, or even countably infinitely many points, you end up with a valid PDF with the same CDF and therefore the same distribution. Anyway, that issue aside, you can view $f(x)$ as describing how relatively likely you are to see a value near $x$. $\endgroup$
    – Ian
    Commented Jun 1, 2016 at 0:43

3 Answers 3

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A Probability Density Function shows the direction of change in the Cumulative Distribution Function at particular points (the gradient, or "slope", of the curve).

Traditionally we once talked about probability in terms of weights of events.   The faces of a fair coin have equal weights, the odds are heavily against a dark horse winning, and such.   The metaphor is still useful.

If a CDF is considered the distribution of accumulated "mass" of probability for a continuous random variable, then its pdf shows how "dense" the accumulation is.   (Hence the names.)

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A pdf does not give probabilities. To explain what a pdf is precisely one would need to utilize measure theoretic concepts. What is important is that we can think of the area under the curve as giving probabilities of being within a particular interval - that is $\int_a^b f(x)dx = \mathbb{P}(a \leq X \leq b)$. The pdf value at any given point is insignificant, and moreover the probability of getting $X=x$ is 0 for all values of $x$.

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One issue is that, unlike probability, probability density is dimensionful—it is probability per unit of some quantity, where that quantity may have units. Changing measurement units will therefore change the probability density. So if $f$ is the pdf for the distribution of heights of adult males, measured in meters, and you suddenly decide to measure heights in centimeters, the corresponding probability density will be $100$ times smaller. This ensures that the probability that height is between $1.6$ m and $1.9$ m is the same as the probability that the height is between $160$ cm and $190$ cm. Specifically, if $g$ is the pdf for heights measured in centimeters, then $g(100x)=0.01\cdot f(x)$, and $$ \int_{160}^{190}g(y)\,dy=\int_{1.6}^{1.9}g(100x)\cdot100\,dx=\int_{1.6}^{1.9}f(x)\,dx. $$ An implication of this is that, by choosing measurement units suitably, you can make probability density have any numerical value you want (including values bigger than $1$).

Because of this dependence on units, probability density is not so fundamental. But that doesn't mean it is meaningless. For example, $f(1.6)$ is approximately the probability that height lies within a one meter interval around $1.6$ m, say $P(1.1\le x\le 2.1)$, while $g(160)$ is approximately the probability that height lies within a one centimeter interval around $160$ cm, say $P(159.5\le y\le 160.5)$. Clearly $g(160)$ is going to be a more reasonable approximation of the quantity it represents than is $f(1.6)$ since the pdf is much closer to being constant between $159.5$ cm and $160.5$ cm than it is between $1.1$ m and $2.1$ m.

The main point is that, in a density function, it is area, not height, that represents probability. The value of $f$ is the height; integrals of $f$ are areas. Values $f(x)$ are useful insofar as they allow you to compute areas. The interpretations of $f(1.6)$ and $g(160)$ in the previous paragraph are based on approximating areas by rectangles (of unit width).

Your relative probability interpretation is basically correct. Your statement that $p(1)/p(0)=4.67$ means that $x=1$ is $4.67$ as likely as $x=0$ would be more correctly stated as $P(1-\epsilon/2\le x\le1+\epsilon/2)$ is approximately $4.67$ times as big is $P(-\epsilon/2\le x\le\epsilon/2)$, or even more correctly as $$ \lim_{\epsilon\to0}\frac{P(1-\epsilon/2\le x\le1+\epsilon/2)}{P(-\epsilon/2\le x\le\epsilon/2)}=4.67. $$ In fact, $P(x=1)$ and $P(x=0)$ are both $0$: for a continuous quantity, the probability that $x$ equals a particular value, to infinite precision, is $0$.

The best way to gain intuition for the pdf is to practice making histograms, which is something you do in a descriptive statistics course. This will give you a feeling for why it is best that area represent probability, and not height. The idea is that, by choosing area, the height and shape of the histogram are relatively insensitive to the choice of bin width, and even to varying the bin widths within a single histogram. For example, if there are $1000$ data points, and $100$ of them lie between $1.6$ m and $1.8$ m then, if heights were used to represent probability, that bin would be a rectangle of width $0.2$ m and height $\frac{100}{1000}=0.1$. Now suppose that $55$ of the data points lay between $1.6$ and $1.7$ and $45$ between $1.7$ and $1.8$. Rebinning would produce two rectangles of width $0.1$ m and heights $0.055$ and $0.045$, which are roughly half the previous height. With areas used to represent probability, the first histogram would have a bin of width $0.2$ m and height $\frac{0.1}{0.2}=0.5\ \mathrm{m}^{-1}$, and the second would have two rectangles of widths $0.1$ m and heights $\frac{0.55}{0.1}=0.55\ \mathrm{m}^{-1}$ and $\frac{0.45}{0.1}=0.45\ \mathrm{m}^{-1}$, which are very close to the original height. By doing things using the area method, you avoid having to change (and reinterpret) your vertical scale every time you change bin size.

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