# Can PMF have value greater than 1?

Can a Probability Mass Function can have value greater than 1 like we have in Continuous functions(pdf) and why? Please give 1 example if the answer is yes. And If the answer is no then why PDF can have value more then 1. As the basic definition of both function is same. F(x):Pr(R=x).

1. Probability Mass Function: It have discrete values and we count only those values for probability. So F(x):Pr(R=x) is only the probabilty which is always less than 1

2. Probability Density Distribution: Here we don't have discrete values and if we consider a point Pr(of single point)=0. Hence we consider area during continous distribution counting. Now since we don't know the exact points(x) for probability distribution there can be cases when probability can be concentrated on a small points rather than the asked points(x). And we also know that area under such PDF curve should be 1. Hence if x is less than 1 then in order to keep area sum 1 value of F(x) should be greater than 1

No, a probability mass function cannot have a value above 1. Quite simply, all the values of the probability mass function must sum to 1. Also, they must be non-negative. From here it follows that, if one of the values exceeded 1, the whole sum would exceed 1. And that is not allowed.

By PMF, I assume you mean what is usually called the pdf. For a continuous distribution the answer is yes. What has to be true is that:

$$\int_{-\infty}^{\infty}p(x) dx = 1$$

Imagine a normal distribution, centered at zero (so a mean of zero), and a timy standard deviation (say, .01). Almost all of the points on that curve that will contribute much to that integral will be between -0.1 and 0.1. So, thinking geometrically and approximating with a rectangle, the height is going to be at least something like 5 (it will actually be a lot bigger, because most of the integral will come from points between -.04 and .04).

The pdf if a non-continuous variable can never be more than 1, since (1) the sum of them must all add to 1 and (2) they must all be non-negative. The largest it can be is if there is only one possible outcome, which then has P(x) = 1.

By the way, the height (greater than 1) has no real meaning in itself, only as part of the integral. It's really tempting for people, when moving from discrete to continuous, to plot out a normal curve with mean = 0 and stdev = 1. If I recall, the high point has a value of around 0.4. But that doesn't mean you will get zero with probability 0.4; in fact, the probability of getting exactly zero (or any other specific number) is zero. But if you change the bounds of the integral to be $x_1$ and $x_2$, you get the probability of getting a number between $x_1$ and $x_2$.

• For discrete distributions it's called a PMF (and that's what the OP is asking about), which looks like a histogram. And they're generally set up so that the sum of the (discrete) bars' heights sum to 1, and under that condition any single bar exceeding 1 would not be permissible. – Josh Aug 18 at 22:09
• Fair enough. When I was taught we always called both a pdf (and CDF) – eSurfsnake Aug 19 at 23:46
• As for CDFs I'm not aware of terms specific to continuous or discrete distributions; they're all called CDFs as far as I know. The difference being, of course, that the CDF of a discrete distribution will have discrete "steps" since it's a sum the discrete probabilities in the PMF, rather than the integral of a continuous PDF. – Josh Aug 20 at 16:16
• Yes, as I was taught one should actually always think of the CDF, which is always consistent: it is 0 and negative infinity, it is 1 at infinity, it is monotonically non-decreasing, and it may have step discontinuities. Then the pdf is its derivative, accounting, of course, for those discrete cases with jumps. Thinking that way helps avoid confusing yourself with questions like 'what does it mean if the pdf has points greater than one?' It means the slope of the CDF was greater than one there, not the actual probability (which is analytically zero at every point).. – eSurfsnake Aug 20 at 19:12