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Probability of continuous random variables

  • #1

Homework Statement


A random variable has distribution function F(z) = P(y<= z) given by (this is a piecewise function)

f(z) =
0 if z < -1
1/2 if -1 <= z < 1
1/2 + 1/4(z-1 if 1 <= z < 2
1 if 2 <= z

What is P(Y = 2)?

Find all the numbers t with the property that both P(Y <= t) >= 1/2 and P(Y >= t) >= 1/2


Homework Equations





The Attempt at a Solution



For the P(Y=2) I integrated at the point 2 plus and minus epsilon and came up with 1/4z - 0 where z =2. Thus, 1/2. My concern is that this problem has a lot of constants Thus, I would expect P(Y=x) to equal 0 everywhere but in [1,2). Then I have no idea how to find the median of the distribution function. Sorry, if these are easy questions. The class is being taught without a book and I'm afraid I'm not used to that.
 

Answers and Replies

  • #2
Hurkyl
Staff Emeritus
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For the P(Y=2) I integrated at the point 2 plus and minus epsilon and came up with 1/4z - 0 where z =2. Thus, 1/2.
I don't understand what you're doing -- could you spell it out? The answer is certainly wrong.

My concern is that this problem has a lot of constants Thus, I would expect P(Y=x) to equal 0 everywhere but in [1,2).
Not just a lot of flat-lines in the graph, but it's mostly continuous! P(Y=a) can only be positive if the graph of the cumulative distribution function has a discontinuity. (right?)

Then I have no idea how to find the median of the distribution function.
Do you know how to find P(Y<=t)?
 
  • #3
MathematicalPhysicist
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well, no expert here, but to find P(Y=2) shouldn't you integrate f(z) over (-infinity,2]?
 
  • #4
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F(z) = P(y<= z)

what exactly is "y" here?

the CDF is supposed to be F(z) = P[Z <= z]
 
  • #5
Hurkyl
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well, no expert here, but to find P(Y=2) shouldn't you integrate f(z) over (-infinity,2]?
He said that f was the cumulative distribution, not the probability density.
 
  • #6
HallsofIvy
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Homework Statement


A random variable has distribution function F(z) = P(y<= z) given by (this is a piecewise function)

f(z) =
0 if z < -1
1/2 if -1 <= z < 1
1/2 + 1/4(z-1 if 1 <= z < 2
1 if 2 <= z
This is impossible. The integral of a probability density function over its entire domain must be 1. You obviously can't have "f(z)= 1 if 2<= z". It also cannot be a cumulative probabililty distribution which was my next guess. I have no idea what is intended here.

well, no expert here, but to find P(Y=2) shouldn't you integrate f(z) over (-infinity,2]?
No, that would give P(Y[itex]\le 2[/itex]). With a continuous probability density, the probobability that Y is any specific number is 0.
 

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