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Determining the distribution function F(x): Statistics/Probability

  • Thread starter mikebro
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  • #1
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Homework Statement



For the random variable X with probability density function determine the distribution function F(x)

http://img20.imageshack.us/img20/3314/questiontsy.jpg [Broken]

Homework Equations



Cumulative Distribution Function


The Attempt at a Solution



Integrate f(x) to get the distribution function:
F(X) =
  • 0, x ≤ -1
  • x[tex]^{2}[/tex]/2 + x, -1 < x ≤ 0
  • x - x[tex]^{2}[/tex]/2, 0 < x ≤ 1
  • 0, x > 1


The actual answer is
http://img297.imageshack.us/img297/426/answer.jpg [Broken]

I understand that the equation must be integrated to get the distribution function, but where do the constants (1/2) come from and how are they calculated?
 
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Answers and Replies

  • #2
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You can't just take a piecewise density function and find the indefinite integral of each piece separately. Notice your F(x) isn't continuous (any integrable density function must have a continuous cumulative distribution function), so it has to be incorrect.

Remember the cumulative distribution function is [itex]\int_{-\infty}^{x}f(t)\,dt,[/itex] where f is your density function (t is just a dummy variable). Now obviously the values of f for x less than or equal to 1 contributes nothing to the cumulative distribution function, so to find the values of F for [itex]-1<x\leq0,[/itex] you need to compute
[tex]\int_{-\infty}^{x}f(t)\,dt = \int_{-\infty}^{x} t + 1\,dt = \int_{-1}^{x} t + 1\,dt. [/tex]
 
  • #3
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Thanks, that helps clarify a bit. I solved the integral you posted and got the correct answer, but tried to solve the other 2 and got an answer different than what the professor gave.

To get the F for [itex]0<x\leq-1,[/itex] I used this integral:
[tex]\int_{-\infty}^{x}f(t)\,dt = \int_{-\infty}^{x} 1 - t\,dt = \int_{0}^{x} 1 - t\,dt. [/tex]

but since F(0) = 0, the answer I got isn't what the professor posted. Does the 1/2 carry over from the F for [itex]-1<x\leq0,[/itex] carry over to this one?

And for the [itex]x>1,[/itex] obviously the F is just 0 + Constant, is it just known that the value of this is 1 since it must add up to one?
 
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  • #4
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3
Does the 1/2 carry over from the F for [itex]-1<x\leq1,[/itex] carry over to this one?
Of course! I mean what you first calculated was the cumulative distribution function for x between -1 and 0, so F(0) = 1/2, not 0. I think you should try drawing the density out to get a geometric feel for the problem for now; this isn't hard once you do a few more of these.

For the last "piece" F(x) should just be 1, but remember you should be able to get this value by computing F(1). The cumulative density function should be continuous, and you should check this at the endpoints of each piece.
 
  • #5
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I actually do have a graph of the density function.
http://img64.imageshack.us/img64/8127/graphbo.png [Broken]

But if F(0) = 1/2, when solving the integral above and doing F(x) - F(0), wouldn't that make it (x^2)/2 + x - (1/2)? What I meant was F(0) = 0 in context of the integral above, then adding the actual value of F(0) from the -1 < x ≤ 0 piece.

I am realizing these seem relatively easy as I work through it, but there's just a few small things I'm unsure of and can't find examples of in my textbook or lecture notes.
 
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