Find cdf, pdf and expextation value of a random variable

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The discussion revolves around finding the cumulative distribution function (CDF), probability density function (PDF), and expected value of a random variable Y defined as Y = X^2, where X has a uniform distribution on the interval [-1, 1]. The initial steps involve determining the CDF F_Y(y) by translating the probability expression from Y back to X, noting that Y only takes values from 0 to 1. The need for proper limits in integration is emphasized to derive the correct CDF formula. Additionally, methods for calculating the expected value E(Y) are discussed, including using the expectation formula or evaluating the characteristic function. The conversation highlights the importance of understanding the behavior of the CDF and proper integration techniques in probability theory.
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Homework Statement



Let X represent the random choice of a real number on the interval [-1,1] which has a uniform distribution such that the probability density function isf_{X}(x)=\frac{1}{2} when -1\leqx\leq1. LetY=X^{2} a. Find the cumulative distribution F_{Y}(y) b. the density function f_{Y}(y) and c. the expected value E(Y).

Homework Equations



my book gives a great explanation on how to change variables for joint distributions, but little is said about functions of one random variable, so I'm kind of at a loss here.

The Attempt at a Solution



first, if Y=X^{2}, then I want to say we need to find Y over the interval
[0,1]. And integrating I have that:
F_{X}(x)=\frac{x+1}{2} which is P(X\leqx)...
now I want to say P(X\leqx)=P(\sqrt{Y}\leqx)...

i'm not sure where to go from here...
can I just substitute \sqrt{y} for x so I have:

F_{Y}(y)=\frac{\sqrt{y}+1}{2} ??
 
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One way to look at the cumulative distribution function F(X) for some random process is F(x) = p(X<x). In other words, the value of the F(x) is the probability that a random number drawn from the distribution is smaller than x. If the random process has some lower bound and upper bound, the CDF F(x) must be identically zero below the lower bound and identically one above the upper bound. Your conjectured function does not exhibit this behavior. (Look at FY(0).)

Using the above definition of the CDF, FY(y) = p(Y<y). Translate this expression to the original random variable and evaluate the resultant probability expression.
 
part a:

1. graph and draw a picture of y = x^2

2. note that y is only valid from 0 to 1 ( since y = x^2 can only produce zero and positive numbers)

3. to get the CDF formula, find the limits of the integral. y goes from 0 to x^2 and x goes from 0 to infinity. once the limits are found, plug the pdf into the CDF formula and then do the integration

part b:

part c:

for part c there are two ways of doing this. use the expectation formula. or find the characteristic function and then evaluate -j times the first derivative of the characteristic function when the characteristic function's variable equals to zero.
 
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Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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