What Is The Pdf Of The Minimum Of The Draw?

PDF is not a probability.

The probability density at x tin can be greater than 1 but then, how can it integrate to one?

Information technology'southward a well-known fact that the largest value a probability can have is 1.

However, for some PDFs (due east.m. the PDF of the exponential distribution, the graph below), when λ= 1.5 and 𝒙 = 0, the probability density is 1.5, which is plainly greater than 1!

From Wikipedia: The PDF of Exponential Distribution

1. How tin a PDF'south value be greater than 1 and its probability still integrate to one?

Even if the PDF f(x) takes on values greater than i, i f the domain that it integrates over is less than ane, it tin can add upward to simply one. Let's accept an example of the easiest PDF — the uniform distribution defined on the domain [0, 0.5]. The PDF of the uniform distribution is 1/(b-a), which is constantly ii throughout.

The total probability is the total area nether the graph f(ten),
which is 2 * 0.5 = 1.

As you can see, even if a PDF is greater than 1, because information technology integrates over the domain that is less than i, it can add together up to 1.

2. The difference between the PDF and probability

Isn't the PDF f(ten) a probability?

No. Because f(ten) can be greater than 1.
("PD" in PDF stands for "Probability Density," non Probability.)

f(𝒙) is only a height of the PDF graph at X = 𝒙. (Are you confused with 𝐗 vs 𝒙 notation? Check it out here.)

The whole "PDF = probability" misconception comes about because we are used to the notion of "PMF = probability", which is, in fact, right. Nevertheless, a PDF is not the same affair every bit a PMF, and it shouldn't exist interpreted in the aforementioned way as a PMF, considering detached random variables and continuous random variables are not defined the aforementioned mode.

For discrete random variables, nosotros wait upward the value of a PMF at a single point to find its probability P(𝐗=𝒙) (eastward.g. Remember how we plugged 𝒙 into the Poisson PMF?)
For continuous random variables, nosotros take an integral of a PDF over a certain interval to detect its probability that X volition fall in that interval.

                              f(ten) ≠ P(X = 𝒙)                            *                  f(x): PDF for a continuous r.five.
                *                  P(Ten = ten) : PMF for a detached r.five.

Sure, all skillful. Nevertheless, you might wonder…

Why do we accept to *integrate* a PDF?

Tin can we just sum upwards PDF values, just similar we practise with PMF values?

No. Because, for continuous random variables, the probability that 𝐗 takes on any particular value 𝒙 is 0.

3. Why is the probability of a continuous random variable at every signal zero?

Let'due south look at the previous example — the Uniform Distribution in [0, 0.five].

The probability density at x=1 is ii.
But why is the probability at x=i cypher?

To answer the above question, we demand to reply the following question kickoff:
How many total real numbers practice we have in [0, 0.five]?

Infinite ∞. (To be mathematically thorough, uncountably infinite.)
0.ane, 0.01, 0.001, 0.0001,… You can keep inserting 0 in forepart of the smallest decimal.

Therefore, a continuous random variable has an infinite # of possible values that it can take, fifty-fifty if the domain is small and fixed. And let'due south say, the probability density for each value in [0, 0.5] takes an extremely small value, e.yard. 000000001. Still, the sum of infinitely (uncountably) many values will reach infinity, no affair how pocket-size their values are.

Then, in order to make the sum of probabilities one, the probability at a unmarried signal should be 1/∞, which is 0.

Well, this doesn't really brand sense, either. If you lot add the infinite number of zeros, you will all the same become cipher. The total probability should add upwards to i, non to zero.

The thing is, we can't use the notion of detached PMF (ane value has one matching probability) for continuous variables. We can't define the probability of continuous variables as we did for detached variables.

4. And so how do we calculate the probability from the probability density f(x)?

Nosotros will borrow the idea from the "integral".

If the probability of X being exactly at point 𝒙 is zero, how about an extremely small interval around the point 𝒙? Say, [𝒙, 𝒙+d𝒙]? Let's say d𝒙 is 0.00000000001.

Then the probability that X will fall in [𝒙, 𝒙+d𝒙] is the area under the curve f(𝒙) sandwiched by [𝒙, 𝒙+d𝒙].

If d𝒙 is infinitesimally small, this approximation volition be practiced enough for P(𝐗=𝒙).

                              f(𝒙)d𝒙                : The probability of                X                in                [𝒙, 𝒙+d𝒙].
                f(𝒙): Probability density.
                d𝒙                : Interval length.

A few things to notation:

1. If you take a await at the definition of both a PDF and a PMF, it's really but changing the summations in the discrete case (PMF) to integrals in the continuous case (PDF).

ii. Why do we employ the terms "Density" and "Mass"?

This is analogous to mass density in physics — integrating the density to get the mass. If you think of a mass as a probability, we are integrating a probability density to get a probability (mass).

3. What does a probability density at point 𝒙 hateful?

It means how much probability is concentrated per unit length (d𝒙) near 𝒙, or how dense the probability is near 𝒙.

4. We need to fix the Wikipedia graph of the exponential distribution. The level of Y-axis P(Ten) sounds similar a probability. We need to alter it to f(x) or "Probability Density".