# Probability mass function and Probability density function

What is the Difference between probability Mass function and probability density function? Why the value of continuous probability distribution function is not the probability for particular input point for example for a continuous distribution say(assume it is pdf under suitable domain). f(x) = x^3 why it is not true that f(3) is probability at 3. what do you mean by f(3)?

Informally, you can think of it as $f(3)$ simply returning the height of the density function, but this is really of no interest. Recall that $f(3)=Pr(X=3)$, and in our case $X$ is a continuous random variable. It does not make sense to speak of probabilities at certain values of $X$ because of this.

For example, let's say I toss a ball. It lands around a 20 feet mark. But, in fact, when I look closer it was more like 19.75. But, looking even closer, it was more like 19.745. But, looking even closer than that, it was more like 19.7445. And so we could continue... It doesn't end! Hence, there is no probability for a certain distance of my toss and we use intervals instead. That is, the probability that I'd toss the ball 20 feet is 0 ($f(20)=Pr(X=20)=0$), but the probability that it's between some interval is not ($Pr(20-\delta < X < 20 +\delta)\neq 0$).

• Really good one... Sep 16, 2013 at 10:58
• please tell me why would we need f(3) if it is not important Sep 16, 2013 at 10:59
• @MilanAmrutJoshi You don't 'need' $f(3)$, but you need $f(x)$ so that you, for example, can calculate the probability that you are in a neighborhood of 3 -- such as $\int_{3-\delta}^{3+\delta}f(x)dx$. It might be helpful using a normal distribution as an example - let $X\sim N(3, 1)$ and $\delta=1.96$. Then $\int_{3-1.96}^{3+1.96}f(x)dx=Pr(3-1.96<X<3+1.96)=Pr(-1.96<X-3<1.96)=0.95$ which should be familiar to you. Choose smaller and smaller values of $\delta$ and see what happens. It's quite well explained in the link, have a look if you haven't already. Sep 16, 2013 at 11:05
• Ok i will have a look .. Sep 16, 2013 at 11:06

I stumbled upon this question and found my answer in MIT OCW notes (Reading 5b and 4a to be precise). I'm quoting the explanation from there.

Probability mass and probability density - these terms are completely analogous to the mass and density you saw in physics and calculus.

Mass as a sum: If masses m1, m2, m3, and m4 are set in a row at positions x1, x2, x3, and x4, then the total mass is m1 + m2 + m3 + m4.

We can define a ‘mass function’ p(x) with p(xj ) = mj for j = 1, 2, 3, 4, and p(x) = 0 otherwise. In this notation the total mass is p(x1) + p(x2) + p(x3) + p(x4). The probability mass function behaves in exactly the same way, except it has the dimension of probability instead of mass.

Mass as an integral of density: Suppose you have a rod of length L meters with varying density f(x) kg/m. (Note the units are mass/length.)

If the density varies continuously, we must find the total mass of the rod by integration:

total mass $$= \int_{0}^{L} f(x) dx$$

This formula comes from dividing the rod into small pieces and ’summing’ up the mass of each piece. That is:

total mass ≈ $$\sum_{i=1}^{n} f(x_i) = \Delta x$$ In the limit as $$\Delta x$$ goes to zero the sum becomes the integral. The probability density function behaves exactly the same way, except it has units of probability/(unit x) instead of kg/m. Indeed, equation (1) is exactly analogous to the above integral for total mass. While we’re on a physics kick, note that for both discrete and continuous random variables, the expected value is simply the center of mass or balance point.

Reference (I couldn't figure out to use "Insert citation" but doing so is required from MIT OCW terms, hence I am manually inserting it here): Jeremy Orloff and Jonathan Bloom. 18.05 Introduction to Probability and Statistics. Spring 2014. Massachusetts Institute of Technology: MIT OpenCourseWare, https://ocw.mit.edu/. License: Creative Commons BY-NC-SA.(Terms : https://ocw.mit.edu/terms/)