Problem with probability theory and random variables

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Homework Help Overview

The discussion revolves around probability theory, specifically focusing on independent random variables and their associated probability distributions. The original poster presents a problem involving the calculation of P(X>Y) for two independent random variables X and Y with the same density function. Another participant introduces a similar problem involving three independent random variables uniformly distributed in the interval [0,1].

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between the probabilities of X and Y, noting that if they have the same distribution, P(XY) to cumulative distribution functions (CDFs) but expresses uncertainty about how to proceed. Others suggest integration as a method to arrive at the probability and question the derivation of certain expressions.

Discussion Status

There is a mix of attempts to clarify the problem and explore different methods of calculation. Some participants provide insights into the integration approach, while others seek clarification on the derivation of specific formulas. The discussion reflects a collaborative effort to understand the underlying concepts without reaching a definitive conclusion.

Contextual Notes

Participants are navigating the complexities of probability distributions and the implications of independence among random variables. The original poster's inquiry is framed within the constraints of homework expectations, which may limit the extent of provided guidance.

trenekas
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Hello. I have a problem with probability theory task.
The task is:
X and Y is independent random variables with same density function fx=fy=f. What will be probability of P(X>Y).

This P(X>Y) reminds me a cdf: P(X>Y)=1-P(X<Y)=1-cdf of X.

Cdf of x is equal to integral ∫f dx from -inf to y.
But don't know how to continue because Y depends on its dencity. Maybe some one will be able to help me?
 
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It should be clear that if x and y have the same probability distribution, then the probability x< y is the same as the probability y< x. And since this is a is a continuous distribution, the probability that x= y is 0.
 
So the answer is 1/2. hm :) ok
 
trenekas said:
So the answer is 1/2. hm :) ok

You can also get it using integration:
P(X&gt;Y) = \int_{-\infty}^{\infty} G(y) f(y)\, dy,
where ##G(y) = P(X>y)##. We have ##f(y) \, dy = -dG(y),## so
P(X&gt;Y) = -\frac{1}{2} \left. G(y)^2 \right|_{-\infty}^{\infty} = 1/2.
 
Also i have very similar question. Suppose that X,Y,Z are independent random variables and uniformly distributed in the interval [0;1]. What will be P(XY<Z^2)
From that i know that density of all thre r.v. is 1 if xε[0,1] and 0 otherwise.
So fx*fy=1 when x[0;1] and 0 otherwise. cdf of this would be F=x when x[0;1] and 0 otherwise...
But what's about Z^2?
 
trenekas said:
Also i have very similar question. Suppose that X,Y,Z are independent random variables and uniformly distributed in the interval [0;1]. What will be P(XY<Z^2)
From that i know that density of all thre r.v. is 1 if xε[0,1] and 0 otherwise.
So fx*fy=1 when x[0;1] and 0 otherwise. cdf of this would be F=x when x[0;1] and 0 otherwise...
But what's about Z^2?

Start by writing down the integrations you need to perform.
 
Ray Vickson said:
You can also get it using integration:
P(X&gt;Y) = \int_{-\infty}^{\infty} G(y) f(y)\, dy,
where ##G(y) = P(X>y)##. We have ##f(y) \, dy = -dG(y),## so
P(X&gt;Y) = -\frac{1}{2} \left. G(y)^2 \right|_{-\infty}^{\infty} = 1/2.
I want to ask, from where you get this:
P(X&gt;Y) = \int_{-\infty}^{\infty} G(y) f(y)\, dy,
G(y) is cdf and f(y) density?
 
trenekas said:
I want to ask, from where you get this:
P(X&gt;Y) = \int_{-\infty}^{\infty} G(y) f(y)\, dy,
G(y) is cdf and f(y) density?
P[X>Y] = ƩP[X>Y|Y\inI]P[Y\inI], where the sum is over a partition of the range of Y. In the limit of vanishingly small intervals I, that produces the integral.
 

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