Relating the CDF to the probability density

In summary: If you integrate by parts using u = x, dv = F_X'(x) \,dx, you should be able to do the integral.In summary, the conversation is about proving that the expected value of a random variable with continuous CDF can be calculated as the integral of the complementary CDF from 0 to infinity. The correct formula is E(X) = \int_0^\infty 1 - F_X(t)\,dt and it can be derived using integration by parts.
  • #1
bonfire09
249
0

Homework Statement


If ##X## is any random variable defined on ##[0,\infty]## with continuous CDF ##F_X(t)##. Prove that ##E(X)=\int_1^\infty (1-F_X(t)) dt##.
.

Homework Equations


The Attempt at a Solution


I am not sure how to go about this. I think double integration can be used to prove it, but I don't want to go down that path. Is there another which I can prove this. Thanks.
 
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  • #2
Are you sure about those limits? I don't see how E(X) can be independent of the behaviour of F between 0 and 1.
 
  • #3
Well it says that ##X## is a random variable with range ##[0, \infty]##.
 
  • #4
bonfire09 said:

Homework Statement


If ##X## is any random variable defined on ##[0,\infty]## with continuous CDF ##F_X(t)##. Prove that ##E(X)=\int_1^\infty (1-F_X(t)) dt##.
.

Homework Equations





The Attempt at a Solution


I am not sure how to go about this. I think double integration can be used to prove it, but I don't want to go down that path. Is there another which I can prove this. Thanks.

Somebody has asked you to prove a false result. The correct version is
[tex] EX = \int_0^{\infty} [1 - F_X(x)] \,dx .[/tex]

Exactly how you should prove it depends on what you know about Lebesgue-Stieltjes integrals, or whether you even need to use them. The proof is easiest if you assume that ##F_X(\cdot)## is absolutely continuous, in which case there is a density ##f_X(\cdot)## on ##(0,\infty)## giving ##F_X(x) = \int_0^x f_X(t) \, dt## for all ##x > 0##. If ##F_X## is not absolutely continuous, it can be continuous but not have a density function; then the proof is harder.

I suggest you try it first for the case where a density function exists. Think integration by parts.
 
Last edited:
  • #5
bonfire09 said:

Homework Statement


If ##X## is any random variable defined on ##[0,\infty]## with continuous CDF ##F_X(t)##. Prove that ##E(X)=\int_1^\infty (1-F_X(t)) dt##.
.


haruspex said:
Are you sure about those limits? I don't see how E(X) can be independent of the behaviour of F between 0 and 1.

I agree; it should be [itex]E(X) = \int_0^\infty 1 - F_X(t)\,dt[/itex].

Homework Equations





The Attempt at a Solution


I am not sure how to go about this. I think double integration can be used to prove it, but I don't want to go down that path. Is there another which I can prove this. Thanks.

This is an application of integration by parts. You want to calculate [tex]
E(X) = \int_0^\infty xF_X'(x) \,dx.[/tex] Note that [tex]
F_X'(x) = -\frac{d}{dx}(1 - F_X(x)).
[/tex]
 

Related to Relating the CDF to the probability density

1. What is the CDF and how does it relate to probability density?

The Cumulative Distribution Function (CDF) is a function that shows the probability of a random variable being less than or equal to a given value. It is used to describe the overall behavior of a probability distribution. The probability density function (PDF) is the derivative of the CDF, and it shows the rate of change of the CDF at a specific point. In other words, the CDF and PDF are closely related, with the CDF representing the accumulation of probabilities and the PDF representing the instantaneous probability at a specific point.

2. How is the CDF calculated?

The CDF is calculated by summing up the probabilities of all values that are less than or equal to a given value. For continuous distributions, it is calculated by integrating the PDF from negative infinity to the given value. The resulting value is between 0 and 1, with 0 representing no probability and 1 representing certainty.

3. What is the difference between discrete and continuous CDFs?

Discrete CDFs are used for discrete random variables, where the possible outcomes are countable and finite. The CDF for discrete variables is a step function, with jumps at each possible value. Continuous CDFs are used for continuous random variables, where the possible outcomes are uncountable and infinite. The CDF for continuous variables is a smooth, continuous function.

4. How is the CDF used in hypothesis testing?

The CDF is used in hypothesis testing to determine the p-value, which is the probability of obtaining a result at least as extreme as the one observed, assuming the null hypothesis is true. The p-value is calculated by finding the area under the curve of the CDF from the observed value to the right (for a one-tailed test) or to both sides (for a two-tailed test). A smaller p-value indicates stronger evidence against the null hypothesis.

5. Can the CDF be used to find the probability of a specific value?

Yes, the probability of a specific value can be found by taking the difference between the CDF at that value and the CDF at the previous value. For example, if the CDF at x=5 is 0.8 and the CDF at x=4 is 0.6, then the probability of x being exactly 5 is 0.8-0.6 = 0.2. However, for continuous distributions, the probability of a specific value is always 0, since the area under a single point on the PDF is 0. Therefore, the CDF is more useful for finding probabilities of ranges of values rather than specific values.

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