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Every night Joe goes to the casino and takes with him an amount of money in dollars, X, that is distributed according to the pdf:

f(x) = Ax^2 for 0 < x < 10

where A is a constant that you need to determine. He returns with an amount of money Y that is uniformly distributed between 0 and 3X. Find

  • E[Y] w/out finding the distribution for Y (i.e. using E[E(Y|X)])
  • The pdf for Y
  • E[Y] using the distribution for Y. Compare with your result of part a.
  • The prob he returns w/ at least 5 dollars
  • If he returns w/ 5 dollars, what is the prob he left with at least 7 dollars
  • If he return w/ 5 dollars, what is the expected value of the amount of money with which he left?

I am having a hard time finding the joint pdf for this situation. I know f(x) = 3x^2/1000 for 0 < x < 10, but I keep getting 1/3X for the distribution of Y, which doesn't seem right. Any hints on how to form the joint pdf here?

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2 Answers 2

Given that $X$ has value $x_0, 0 < x_0 < 10$, $Y$ is (conditionally) uniformly distributed on $[0, 3x_0]$ and hence $E[Y\mid X=x_0] = 3x_0/2$. (If you can't do this last step by inspection, work out the value of $\int_0^{3x_0} y \cdot \frac{1}{3x_0} \,\mathrm dy$). Since $E[Y\mid X=x_0]$ depends on the value of $X$, it can be regarded as a random variable_ that is a function of the random variable $X$). We denote this random variable by $E[Y\mid X]$ and note that it happens to be the function $3X/2$ in this instance. Can you now find $E[Y] = E[E[Y\mid X]] = E[3X/2]$ without needing to find the pdf of $Y$?

For the pdf of $Y$, note that $$f_{Y\mid X}(y\mid x) = \begin{cases}\frac{1}{3x}, &0 < y < 3x,\\\quad\\0, &\text{otherwise,}\end{cases}$$ and so, $$f_{X,Y}(x,y) = f_{Y\mid X}(y\mid x)f_X(x) = \begin{cases}\frac{1}{3x}\cdot\frac{3x^2}{1000}, &0 < y < 3x, 0 < x < 10,\\\quad\\0, &\text{otherwise.}\end{cases}$$ You can find the pdf of $Y$ from this. Remember to sketch the region where the joint pdf is nonzero; it will help you set the limits of integration correctly.

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I'm attempting to solve for E[Y] using the formula E[Y] = integral(E[Y|X]f(x)dx) from 0 to 10. Would this give the proper E[Y]? I got 45/4 for example. –  sixglazed May 23 '13 at 1:21
    
@sixglazed Yes, $E[Y] = \frac{45}{4}$. –  Dilip Sarwate May 23 '13 at 2:49

The conditional density of Y given X=x is 1/(3x). You need to find the unconditional distribution by integrating over f(x).

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Could you guide me in how to set up this integral? What would f(x) be in this case and what bounds would you use? –  sixglazed May 23 '13 at 0:03
    
f(x) would be the PDF of x and the limits would be 0 to 10. –  response May 23 '13 at 0:12
    
Isn't the integral of 3x^2/1000 from 0 to 10 just 1? –  sixglazed May 23 '13 at 0:16
    
You integrate f(x) 1/(3x). –  response May 23 '13 at 0:56

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