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I understand that the probability mass function of a discrete random-variable X is $y=g(x)$. This means $P(X=x_0) = g(x_0)$.

Now, a probability density function of of a continuous random variable X is $y=f(x)$. Wikipedia defines this function $y$ to mean

In probability theory, a probability density function (pdf), or density of a continuous random variable, is a function that describes the relative likelihood for this random variable to take on a given value.

I am confused about the meaning of 'relative likelihood' because it certainly does not mean probability! The probability $P(X<x_0)$ is given by some integral of the pdf.

So what does $f(x_0)$ indicate? It gives a real number, but isn't the relative likelihood of a specific value for a CRV always zero?

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Let $f$ be the density function of $X$. Assume $f$ is continuous. Then if $h$ is small, the probability that $X$ lies in the interval $[a,a+h]$ is approximately $hf(a)$. By approximately I mean that the probability, divided by $h$, approaches $f(a)$ as $h$ approaches $0$. So the ratio $f(a)/f(b)$ measures, approximately, the ratio of the probability that $X$ is in $[a,a+h]$ to the probability $X$ is in $[b,b+h]$. –  André Nicolas Oct 10 '12 at 19:28

2 Answers 2

up vote 2 down vote accepted

'Relative likelihood' is indeed misleading. Look at it as a limit instead: $$ f(x)=\lim_{h \to 0}\frac{F(x+h)-F(x)}{h} $$ where $F(x) = P(X \leq x)$

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So you suggest looking at the pdf as being defined by the cumulative distribution function? –  jesterII Oct 10 '12 at 19:39
This is essentially the definition of pdf fro CRVs –  Alex Oct 10 '12 at 20:12
A good way of thinking about is $f(x) = \frac{dF}{dx}$ and so it's the rate of change of the cdf at $x$. –  Jacob Feb 27 '13 at 17:39
Why the downvote? What's wrong with what I wrote? –  Alex Feb 27 '13 at 17:48

In general, if $X$ is a random variable with values of a measure space $(A,\mathcal A,\mu)$ and with pdf $f:A\to [0,1]$, then for all measurable set $S\in\mathcal A$, $$P(X\in S) = \int_S fd\mu $$ So, if $A=\Bbb R$ (and $\mu=\lambda$), then $$P(a<X<b)=\int_a^b f(x)dx$$ So, $f(x) = \displaystyle\lim_{t\to 0} \frac1{2t}\int_{x-t}^{x+t} f =\lim_{t\to 0} \frac1{2t} P(|X-x|<t) $ for example.. We can call it 'relative likelihood'..

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This is a better answer than Alex's but doesn't explain the significance of the number $f(x)$. Does it have a meaning independent of a cdf? Andre's answer of it being approximately $hf(a)$ is great but he doesn't indicate if there's more to $f(x)$ by itself. –  Jacob Feb 27 '13 at 17:23

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