(Easy?) Probability Q: Uniform in Uniform

In summary, you are struggling with a probability question and you need help with the expectation equation.
  • #1
ryzeg
9
0
Hello

I have been struggling with a simple probability question.

Homework Statement



We are given that X is a uniformly distributed random variable on [0, 1]. After X is chosen, we take another uniform [0, 1] random variable Y (independent of X) and choose the subinterval that Y falls in. L is the length of this chosen subinterval (either [0, X] or (X, 1]), and we want its expectation.

Homework Equations



The expectation of a uniform random variable is (a+b)/2, where a and b are the interval bounds. These equations may also be useful:

[tex]f_{Y}(y|X=x) = \frac{f(x,y)}{f_{X}(x)}[/tex]

[tex]f_{X}(x|Y=y) = \frac{f_{Y}(y|X=x)f_{X}(x)}{f_{Y}(y)}[/tex]

[tex]E(g(Y)|X=x) = \int \! g(y)f_{Y}(y|X=x) \,dx[/tex]

[tex]E(Y) = \int \! E(Y|X=x)f_{X}(x) \,dx[/tex]

The Attempt at a Solution



I notice

[tex]L=\left\{\begin{array}{cc}X,&\mbox{ if }
Y\leq X\\1-X, & \mbox{ if } Y>Z\end{array}\right[/tex]

And E(X) is obviously 1/2... but I do not know how to find E(L) :(

I tried doing double integrals for expectation on both cases, and summing these, but I get 1/2 which doesn't seem right...

[tex]E_0(L) = \int_{0}^{1} \int_{0}^{x} \! 1 y \, dy \,dx = 1/3[/tex]

[tex]E_1(L) = \int_{0}^{1} \int_{x}^{1} \! 1 y \, dy \,dx = 1/6 [/tex]

[tex]E(L) = E_0 + E_1 = 1/3 + 1/6 = 1/2[/tex]

Any help?
 
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  • #2
Where is the "y" in your expectation integrals coming from?
The value function is either f(x) = x (when y < x) or f(x) = 1 - x (when y > x). I don't see that in your integrals.

Recall that [tex] \int f(x)g(x) dx [/tex] is the integral for expectation where f(x) is the value function and g(x) is the density. Make sure when you construct your integrals that you have the right values and densities.
 
  • #3
I was using y as the value function, with the bounds ranging over all possible values.

When I do integrals with x*1 from 0 to 1, I still get 1/2.
 
  • #4
Are these your integrals?

[tex]
E_0(L) = \int_{0}^{1} \int_{0}^{x} x \, dy \,dx
[/tex]

[tex]
E_1(L) = \int_{0}^{1} \int_{x}^{1} 1 - x \, dy \,dx
[/tex]

They don't add up to 1/2.
 
  • #5
Yeah, you're right, I did my integrals wrong. By symmetry, it is obvious that it should be 2/3, which is what those add up to. Thanks!
 
  • #6
hi there I am new with this stuff but need some help... I've given a question in Finding C,D, F(x). Where the mean is 10 standard deviation is 1. (Uniform Probability distribution) Can you tell me which formula i should used...
Thanks
 

Related to (Easy?) Probability Q: Uniform in Uniform

What is the concept of "uniform in uniform" probability?

"Uniform in uniform" probability refers to a type of probability distribution where all outcomes have an equal chance of occurring. In other words, the probability is evenly distributed across all possible outcomes.

How is "uniform in uniform" probability different from other types of probability distributions?

Unlike other probability distributions, such as the normal distribution or binomial distribution, the "uniform in uniform" probability distribution does not favor any particular outcome. This means that each outcome has the same probability of occurring, making it a special case of the discrete uniform distribution.

What are some real-life examples of "uniform in uniform" probability?

One example of "uniform in uniform" probability is a coin toss, where the probability of getting heads or tails is equal. Another example is rolling a fair die, where each of the six numbers has an equal probability of being rolled.

How is "uniform in uniform" probability calculated?

The formula for calculating "uniform in uniform" probability is P(x) = 1/n, where n is the total number of possible outcomes. This means that the probability of any given outcome is 1 divided by the total number of outcomes.

What are some common misconceptions about "uniform in uniform" probability?

One common misconception is that all probabilities are "uniform in uniform." In reality, there are many different types of probability distributions, and "uniform in uniform" is just one of them. Another misconception is that outcomes are equally likely in every scenario, which may not always be the case as external factors can influence the probability distribution.

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