# Math Help - probability density functions

1. ## probability density functions

If X is an continuos random variable as there are infinitel many elements which belongs to X the $P(X=a)$ cannot be calculated,

But if it is given that $p(X=x) = f(x)$, when $x=a$we can get $P(X=a)$

How come?

If X is an continuos random variable as there are infinitel many elements which belongs to X the $P(X=a)$ cannot be calculated,

But if it is given that $p(X=x) = f(x)$, when $x=a$we can get $P(X=a)$

How come?
Please post the original question that led you to ask this one.

3. sorry for the late reply,

it wasn't a question that led me to ask this one, it was something that my lecturer said

He told that in a cumulative density function we get $p(X \leq x)$from the $F(x)$ which we can also equal to $p(X < x)$

if this is true, since

$P(X=x) = P(X \leq x) - P(X < x)$

$P(X) = 0$

but according to the pdf, we get a value for $P(X=x)$ how can this happen?

it wasn't a question that led me to ask this one, it was something that my lecturer said

He told that in a cumulative density function we get $p(X \leq x)$from the $F(x)$ which we can also equal to $p(X < x)$

if this is true, since

$P(X=x) = P(X \leq x) - P(X < x)$

$P(X) = 0$

but according to the pdf, we get a value for $P(X=x)$ how can this happen?
The pdf can have a non-zero value but the integral of the pdf is equal to zero. And it is the integral that's used to calculate a probability.

5. then what do we get from the pdf?

6. $f_{X}(x)$,the pdf, is the density of random variable X evaluated at x.

7. sorry still confused...

When you graph the pdf function, the area under the graph indicates the interval under which the variable falls.

9. The probability density function is just that - it is a measure of the density of some random variable; or rather "how much" of that random variable there is compared to the whole. Think back to probability histograms - the probability that some value was equal to one of your observed values was the number of values in your population that have that value, divided by the total number of values you had. The number of values you had was the values "density" - how much of the total there is.

Here we are dealing with continuous random variables, and the densities are no longer discrete, but defined by some function f(x). So, if you plug in a number X=x - the density function will tell you how much/many of "X" you have. It WONT tell you the probability of "X" however. To do that you need to find out how much AREA (or volume, or - shudder - hypervolume) your random variable(s) takes on - compared to the total measure, which - as you know - is always 1.

The easiest pdf to work with in helping to motivate understanding is the uniform distribution on (0,1). You should verify that the pdf (density) of ANY random variable is 1. Thus f(0.3857362)=1. The density of 0.3857362 is 1. Now, the confusing part - I'm guessing - comes from distinguishing between density and probability. There are 1 of 0.3857362 - so you'd think - well ok lets just count how many random variables we have and divide by that number to get the probability of getting 0.3857362. Obviously we can't do that - since there are infinite number of random variables (and also why the probability of observing that SPECIFIC value is pretty much zero). That's why we take an interval: (a,b) for example. I can then calculate the area that the interval (a,b) takes up (or volume if I'm dealing with a joint pdf), and divide it by the TOTAL area that is taken up - which will always be 1. And how do you calculate area - integration. That is why probability of observing an interval (a,b) is found by integrating the density function.

Hopefully this helps a little bit in pushing you to delineate between the two and why we need the density function.

10. so what you're telling is that if you want to get the probability of happening x, f(x) will not give the answer?