Expected value of X and Y, E[XY] for uniform random variables

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The discussion focuses on determining the covariance between two random variables, X and Y, where X is uniformly distributed over (-1, 1) and Y is defined as X squared. The covariance formula is presented, leading to the conclusion that cov(X, Y) simplifies to E[XY] since E[X] is zero. The participants suggest substituting Y with X^2 in the expectation calculation, emphasizing that Y is effectively defined as X^2. This relationship allows for a clearer understanding of how to compute E[XY]. The thread concludes that since Y is directly derived from X, calculating E[XY] becomes straightforward.
Dustinsfl
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Homework Statement


If ##X\sim\mathcal{U}(-1,1)## and ##Y = X^2##, is it possible to determine to ##cov(X, Y)##?

Homework Equations


\begin{align}
f_x &=
\begin{cases}
1/2, & -1<x<1\\
0, & \text{otherwise}
\end{cases}\\
f_y &=
\begin{cases}
1/\sqrt{y}, & 0<x<1\\
0, & \text{otherwise}
\end{cases}
\end{align}

The Attempt at a Solution


$$
cov(X,Y) = E[XY] - E[X]E[Y] = E[XY] - 0\cdot 1/2 = E[XY]
$$
Now
$$
E[XY] = \int_0^1\int_{-1}^1g(X, Y)f_{x,y}(x,y)dxdy
$$
From the information that I have, can I determine ##E[XY]##?
 
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I suggest using the fact that ##Y = X^2##...
 
Orodruin said:
I suggest using the fact that ##Y = X^2##...

How?
 
What do you get if you replace ##Y## everywhere by ##X^2##?
 
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Dustinsfl said:
How?

For any observed value ##X = x##, the observed (or, rather, computed) value of ##Y## is ##y = x^2##. It not just that ##Y## and ##X^2## have the same distribution; much more than that is true: ##Y## IS ##X^2##.
 
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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