Expectation and Variance for Continous Uniform RV

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Homework Statement



8. Suppose that X and Y are independent continuous random variables, and each is uniformly distributed on the interval [0,1] (thus the pdfs for X and Y are zero outside of this interval and equal to one on [0,1]).

(a) Find the mean and variance for X+Y.

(b) Calculate and graph the pdf for X+Y (using the convolution formula, and be careful to remember that the pdf of X vanishes outside of [0,1], etc).

Homework Equations



fX+Y = fX * fY

X + Y = Z


The Attempt at a Solution



I know that the density functions for X and Y are both 1 since the continuous distribution is 1/(b-a). To get the density function for Z, I calculated [tex]\int[/tex] fX(x) * fY(z-x) dx from 0 to z. Since this simplifies to [tex]\int[/tex] 1 dx from 0 to z. I get fZ(z) = z.

Using the density function of Z, I calculate the mean, which is E(Z = z) = [tex]\int[/tex] z2 dz from 0 to 1. I get 1/3 for my mean when I evaluate the integral.

To get the second moment, i.e. E(Z2) : [tex]\int[/tex] z3 dz from 0 to 1. When I evaluate this I get 1/4. To get the variance I then get E(Z2) - E(Z)2 which is 1/4 - 1/9 = 5/36.

I don't know if this is right though.
 

Answers and Replies

  • #2
LCKurtz
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Remember that X and Y take on values between 0 and 1. So X+Y can take on values between 0 and 2. Draw some lines for z = x+y = various constants on your square. If you calculate [itex] P(Z\le z)[/itex] directly from your uniform distribution on the square, you will probably see what the instructors caveat about the densities being 0 off the square has to do with it.
 
  • #3
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I calculated the mean of X+Y to be 1. The variance resulted in 1/6. I am not getting the same answer when I use the density function for Z that I tried calculating. Even when I use the limits from 0 to 2, I get a different mean and variance is negative which tells me I am wrong somewhere.

I'm quite sure I did part a correctly, but I am still struggling with part b.
 
  • #4
LCKurtz
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I calculated the mean of X+Y to be 1. The variance resulted in 1/6. I am not getting the same answer when I use the density function for Z that I tried calculating. Even when I use the limits from 0 to 2, I get a different mean and variance is negative which tells me I am wrong somewhere.

I'm quite sure I did part a correctly, but I am still struggling with part b.

No, you don't have part a correct. Just check it:

[tex]\hbox{Does }\int_0^2 f_Z(z) dz = 1\hbox{ ?}[/tex]

If you show your calculations it might be possible to point out your mistake. If you are using the convolution formulas, are you starting something like this, using the convolution formula:

[tex]f_Z(z) = \int_{-\infty}^{\infty}f_X(x)f_Y(z-x)dx = \int_{0}^{1}1\cdot f_Y(z-x)dx [/tex]

If so, you need to take care about where the values of z lie and the value of fY
 
  • #5
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Here is my work for part a. I got the answer without finding the pdf for X+Y. When I tried using pdf for X+Y which I called Z, the answers weren't matching. [PLAIN]http://img338.imageshack.us/img338/1226/partaprob.jpg [Broken]
 
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  • #6
LCKurtz
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Here is my work for part a. I got the answer without finding the pdf for X+Y. When I tried using pdf for X+Y which I called Z, the answers weren't matching.

That is because you haven't calculated the pdf of X+Y correctly. I have given you two different suggestions how to do that.
 

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