Random Variables Transformation

In summary, the problem involves finding the probability that the product of two independent random variables, X and Y, is greater than the square of another independent random variable Z. Using the fact that all variables have a uniform distribution in the range (0,1), it is possible to solve this problem without using the joint probability function. The solution involves setting up inequalities and using the cumulative distribution function of Z to find the desired probability.
  • #1
libelec
176
0

Homework Statement



X, Y, Z random variables, independent of each other, with uniform distribution in (0,1). C = XY and R = Z2. Without using the joint probability function, find P(C>R).

The Attempt at a Solution



So far:

P(C > R) = P(C - R > 0) = P(XY - Z2 > 0) = P(g(X,Y,Z) > 0).

Now, I don't know how to find g-1(-infinity, 0) = {(x,y,z) in (0,1)3 : xy - z2 > 0} to continue to operate. Or any other way to solve this.

Any suggestions?
 
Physics news on Phys.org
  • #2
Try [itex]-\sqrt{XY} < Z < \sqrt{XY}[/itex] instead, and remember that all variables are uniform over (0,1).
 
  • #3
Let me see.

P(C > R) = P(Z2<XY) = P([tex] - \sqrt {XY} < Z < \sqrt {XY} [/tex]) = FZ([tex]\sqrt {XY} [/tex]) - FZ(-[tex]\sqrt {XY} [/tex]), where in the second equality I used that all values of XY are positive and in the last one that FZ is absolutely continuous.

But now, since Z ~ U(0,1), its FZ(z) = z 1{0<z<1} + 1{z>1}. So FZ(-[tex]\sqrt {XY} [/tex]) = 0, because FZ(z) is 0 for all z<0.

So, I would get that P(C > R) = FZ([tex]\sqrt {XY} [/tex]) = sqrt(xy) 1{0<sqrt(xy)<1} + 1{sqrt(xy)>1}. XY is bounded by 1, so sqrt(xy) can't be >1. Then this is sqrt(xy) 1{0<sqrt(xy)<1}.

I'm sorry, but I don't know how to continue...
 
  • #4
And now what do I do? How can I use that X and Y have uniform distributions?
 

FAQ: Random Variables Transformation

What is a random variable transformation?

A random variable transformation is a mathematical process that changes the values of a random variable to a new set of values. This is often done to simplify the calculation of probabilities or to make the data easier to interpret.

Why is random variable transformation important in statistics?

Random variable transformation is important in statistics because it allows for the analysis of complex data sets and the calculation of probabilities for non-standard distributions. It also helps to identify relationships between variables and can improve the accuracy of statistical models.

What are some common types of random variable transformations?

Some common types of random variable transformations include logarithmic, exponential, and power transformations. Other common transformations include square root, inverse, and Box-Cox transformations.

How do you determine which random variable transformation to use?

The type of random variable transformation to use depends on the distribution of the data. To determine which transformation to use, you can plot the data and check for patterns or use statistical tests to identify the best transformation for your data.

What are the potential drawbacks of random variable transformations?

One potential drawback of random variable transformations is that they can sometimes distort the original data and make it difficult to interpret. Additionally, choosing the wrong transformation can lead to incorrect conclusions and biased results. It is important to carefully consider the data and choose the appropriate transformation method.

Back
Top