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Let X and Z be independent random variables with X uniformly distributed on (−1, 1) and Z uniformly distributed on (0, 0.1). Let $Y = X^2 + Z$. Then X and Y are dependent.

  1. Find the joint pdf of X and Y.
  2. Find the covariance and the correlation of X and Y.

I am kind of confused about how to do this. I have been looking through my book for a similar problem or example but can not find it. I just need help with (1) the other part I can figure out pretty simply if I know the joint pdf of X and Y.

I thought of doing this but got nowhere fast $$f_X(x)=\int f_{XY}(x,y)dy$$ and trying to solve for $f_{XY}(x,y)$.

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1 Answer 1

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$Y=X^2+Z$ means $Z=Y-X^2$, so we let $z(x,y)=y-x^2$

After affirming that there is a bijection between $(X,Z)\leftrightarrow(X,Y)$ , (because why?), we then can directly apply the Jacobian change of variables transformation:

$$\begin{align}f_{X,Y}(x, y) =&~ f_{X,Z}(x, z(x,y))~\big/\Big\lvert\dfrac{\partial\big(x,z(x,y)\big)}{\partial\big(x,y\big)}\Big\rvert \\[1ex] =&~ f_{X,Z}(x, y-x^2)~\Big/\begin{Vmatrix}\partial x/\partial x & \partial x/\partial y\\[1ex] \partial (y-x^2)/\partial x & \partial (y-x^2)/\partial y \end{Vmatrix} \end{align}$$

Can you complete?

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  • $\begingroup$ Thanks for the help. Now from this I believe I figured it $f_{XY}(x,y)=5$ for $-1 \le x \le 1$ and $x^2 < y <x^2 +0.1$. I noticed that there is some typos in your solution. It should be the inverse of the jacobian and absolute value : $$\left|\begin{matrix} \frac{\partial x}{\partial x} & \frac{\partial x}{\partial y}\\ \frac{\partial z}{\partial x} & \frac{\partial z}{\partial y} \end{matrix}\right|^{-1}$$ $\endgroup$
    – Andrew
    Jul 21, 2016 at 0:13

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