Finding Density for V=(2Y1+1)^2 on U(-1,1)

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In summary, the problem involves finding the density of the random variable V, which is equal to (2Y1+1)^2, given that Y1 is a random sample from a U(-1,1) distribution. The method of transformations for random variables is used to solve this problem. However, there may be an error in the solution as the range of 2Y1+1 should be from -1 to 3, not 1 to 3.
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James08
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Homework Statement



Suppose {Y1,...Yn} are random samples from U(-1,1) distribution. Find the density of V=(2Y1+1)^2

Homework Equations



Method of transformations for random variables

The Attempt at a Solution



I started this problem off by letting X=2Y1+1 because i figured i could find the density of that and then just find the density of V=X^2. I thought i had the problem right but when i went back to the density i got for X=2Y1+1 and integrated it over the interval, it didnt add up to 1. Not sure what i did wrong
 

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  • #2
Be careful about the range of the random variable. As Y1 ranges from -1 to +1, what is the range of 2Y1+1?

RGV
 
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  • #3
Ray Vickson said:
Be careful about the range of the random variable. As Y1 ranges from -1 to +1, what is the range of 2Y1+1?

RGV

-1 to 3 right?
 

1. What is density in the context of U(-1,1)?

Density in this context refers to the distribution of values within the interval between -1 and 1. It describes how likely it is for a randomly chosen number to fall within a specific range within that interval.

2. How is density calculated for a uniform distribution?

The density for a uniform distribution is calculated by dividing 1 by the length of the interval. In this case, the interval is 2 (1-(-1)=2), so the density would be 1/2 or 0.5.

3. Why is it important to find density on U(-1,1)?

Finding density on U(-1,1) is important in order to understand the probability of a certain value occurring within that interval. This information can be used in various statistical analyses and simulations to make more accurate predictions.

4. How does the density change if the interval is expanded or contracted?

The density will change proportionally if the interval is expanded or contracted. For example, if the interval is expanded to U(-2,2), the density would be 1/4 or 0.25. If the interval is contracted to U(-0.5,0.5), the density would be 1 or 1.

5. Can density be greater than 1?

No, density cannot be greater than 1. It describes the likelihood of a value occurring within an interval, and if the density is greater than 1, it would imply that there is a higher chance of a value occurring than the entire interval itself, which is not possible.

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