P(X=x) in continuous distributions

In summary, the conversation discusses the idea of probability for continuous distributions and how it relates to the exponential pdf. While a continuous random variable has a probability of zero of equaling any specific value, it is still possible to draw conclusions about related probabilities given a particular value. The conversation also touches on the incorrect way of thinking about an integral as an infinite sum and how it can lead to wrong conclusions. It is important to use proper mathematical reasoning when dealing with probabilities and integrals.
  • #1
mertcan
340
6
upload_2016-10-26_18-59-27.png

hi,initially I am aware that for continuous distributions, P(X=x) always equals zero, but when I look at some derivations as the attachment I see that for exponential variable they use exponential pdf when they want to find P(X1=x). My question is : if we say that for continuous distributions, P(X=x) always equals zero why P(X1=x) equals exponential pdf here why it is not 0 ?
 
Last edited:
  • Like
Likes Stephen Tashi
Physics news on Phys.org
  • #2
There is no P(X1=x) in the calculations. There are probabilities of other things given X1=x. That is different. Although the probability of a continuous random variable equaling any pre-specified value is zero, a sample of the variable must have a value, say x0. Given that the value of X is x0, it is legitimate to draw conclusions regarding other related probabilities.
As a simple example, suppose X is a uniform random variable in [0, 10]. Define the related random variable Y = 0 if X<1; Y=1 if X ≥ 1. Then even though P( X = π ) = 0, we can legitimately say that P( Y = 1 | X = π ) =1.
 
Last edited:
  • #3
mertcan said:
if we say that for continuous distributions, P(X=x) always equals zero why P(X1=x) equals exponential pdf here why it is not 0 ?

In doing calculations, it often possible to think informally of the a continuous density function f(x) as giving a probability P(X = x) even though it does not. This invalid way of thinking is helpful in remembering certain formulas - even though it doesn't prove them.The derivation you showed didn't quote any theorems to justify its steps. So let's ask how your text justified the first step:

##P\{ X_1 < X_2\} = \int_{0}^\infty P\{X_1 < X_2| X_1 = x \} \lambda_1 e^{-\lambda_1 x} dx##

If the author of your text is very careful, you will be able to find a theorem or definition that justifies this step. If the author of your text is giving an informal treatment of probability then he may expect the reader to justify the first step by thinking:

##P\{ X_1 < X_2\} = \int_{0}^\infty P\{X_1 < X_2| X_1 = x \} P\{X_1=x\} dx##
## = \int_{0}^\infty P\{X_1 < X_2| X_1 = x \}\lambda_1 e^{-\lambda_1 x} dx##

This informal way of thinking applies several ideas. One idea is that the probability of an event can be computed by partitioning it into mutually exclusive events and adding up the probabilities of the events in the partition. So we partition the event ##\{X_1 < X_2\}## by considering all possible values of ##X_1##. Informally , we are thinking of the set ##\{X_1 < X_2\}## as a union of sets like:

##\{X_1 < X_2\} =( \{X_1 < X_2\}\cap\{X_1 = 0.3\}) \cup ( \{X_1 < X_2\}\cap\{ X_1 = 0.62\} ) \cup (\{X_1 < X_2\} \cap \{X_1 = 5.7\}) \cup ...##

##P\{X_1 < X_2\} = P ( \{X_1 < X_2\}\cap\{X_1 = 0.3\}) + P( \{X_1 < X_2\}\cap\{ X_1 = 0.62\}) + P (\{X_1 < X_2\} \cap \{X_1 = 5.7\} ) + ...##

( Since there are as many possible values of ##X_1## to consider as there are non-negative real numbers, we can't effectively express these thoughts by using a notation with "##...##". )

The next idea is applying the theorem ##P\{A \cap B\} = P\{A | B\} P\{B\} ## to each term in the sum, obtaining:

##P\{X_1 < X_2\} = P \{X_1 < X_2| X_1 = 0.3\}P\{X_1 = 0.3\} + P \{X_1 < X_2| X_1 = 0.62\}P\{ X_1 = 0.62\} + P \{X_1 < X_2| X_1 = 5.7\} P\{ X_1 = 5.7\} + ...##

Next we use the hazy notion that "An integral is an infinite sum", obtaining

##P\{X_1 < X_2\} = \int_0^\infty P\{X_1 < X_2| X_1 = x\} P\{X_1 = x\} dx##

Using the misinterpretation ##P \{X_1 = x\} = \lambda_1 e^{-\lambda_1 x} ## we get the first step shown in your text:
## P\{X_1 < X_2\}= \int_{0}^\infty P\{X_1 < X_2| X_1 = x \}\lambda_1 e^{-\lambda_1 x} dx##

This way of thinking is analgous to the type of play known as a "Comedy of Errors". Many mishaps lead to a happy conclusion.

It is valid to think of a set as being partitioned into as many subsets as there are real numbers. Instead of a "##...##" notation we should use notation like:

##\{X1 < X2\} = \cup_{x \in [0,\infty)} ( \{X_1 < X_2\} \cap \{X_1 = x \}) ##

However, it is not valid to apply the idea that "The probability of an event is the sum of the probabilities of the mutually exclusive events that partition it" to such a partition. When we study formal probability theory in terms of "measure theory" we use axioms that apply this idea only to certain partitions composed of a countably infinite or finite number of subsets. (For example, the length of [0,1] is not the sum of the "lengths" of each point in [0,1]. But we can say the length of [0,1] = length of [0,1/2) + length of [1/2, 3/4) + length of [3/4, 7/8) + ... .)

The hazy idea that "An integral is an infinite sum" doesn't do justice to the concept of an integral. An integral can be viewed as a limit of sums, but they are not simply sums of the values of a function. For example ##\int_0^1 f(x) dx## isn't ## lim_{n\rightarrow \infty} \sum_{i = 0}^n f( i/n) ##. A sum involved in an integral involves the values of the function multiplied by another factor - e.g. ##\sum_{i=0}^n f(i/n) (1/n) ##.

If we wanted to straighten out the wrong steps in the above comedy of errors, we could improve them by using the type of "##dx,##" reasoning used in physics. That type of reasoning does not give a rigorous proof, but it is fairly reliable in deducing correct conclusions. If we applied that type of reasoning, you would find that we are only dealing with probabilities like ## P\{ x - dx/2 < X_1 \leq x + dx/2 \} ## and we are approximating such a probability by ##\lambda_1 e^{-\lambda_1 x} dx ##. We would not be considering the event ##P\{X_1 = x\} ##.

It would interesting to perform the proof in your text using the "##dx##" type of reasoning - but it's late at night and I still have other things I need to do!
 
  • Like
Likes mertcan
  • #4
@Stephen Tashi thanks for your remarkable and nice answer, I so glad , but I am really eager to examine the derivation, I hope you can provide me with the proof of "dx" type reasoning if you have plenty of time...
 

What is the definition of P(X=x) in continuous distributions?

P(X=x) in continuous distributions refers to the probability that a continuous random variable X takes on a specific value x. In other words, it is the likelihood that X will fall within a particular range of values around x.

How is P(X=x) calculated for continuous distributions?

P(X=x) is calculated by finding the area under the probability density curve for the continuous distribution at the specific value x. This is done using calculus and is represented by the integral of the probability density function.

Can P(X=x) be greater than 1 in continuous distributions?

No, P(X=x) cannot be greater than 1 in continuous distributions. This is because the total area under the probability density curve is equal to 1, representing the total probability of all possible outcomes.

Can P(X=x) be negative in continuous distributions?

No, P(X=x) cannot be negative in continuous distributions. This is because probability values cannot be negative and the area under the probability density curve must always be positive.

How does P(X=x) differ from P(X≤x) and P(X≥x) in continuous distributions?

P(X=x) refers to the probability of a specific value, while P(X≤x) and P(X≥x) represent the probability of X being less than or equal to x, and greater than or equal to x, respectively. These probabilities are calculated by finding the area under the probability density curve for the specified range of values.

Similar threads

  • Set Theory, Logic, Probability, Statistics
Replies
1
Views
787
  • Set Theory, Logic, Probability, Statistics
Replies
5
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
5
Views
852
  • Set Theory, Logic, Probability, Statistics
Replies
2
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
2
Views
987
  • Set Theory, Logic, Probability, Statistics
Replies
1
Views
4K
  • Set Theory, Logic, Probability, Statistics
Replies
8
Views
648
  • Set Theory, Logic, Probability, Statistics
Replies
4
Views
990
  • Set Theory, Logic, Probability, Statistics
Replies
3
Views
2K
Replies
0
Views
234
Back
Top