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Integral and probability

  • Thread starter Cosmossos
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  • #1
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Homework Statement


The probability is given by:
p(x,y)=6(1-x-y)
x>=0
y>=0
x+y<=1

Show that p is normilized.
calculate <x>


Homework Equations


I found that p is normilized using the integral:
untitled.JPG

and It's equal to 1.
are the limits of the integral correct?
 

Answers and Replies

  • #2
gabbagabbahey
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Why are you integrating [itex]y[/itex] from [itex]x[/itex] to [itex]1-x[/itex]? Are you told that [itex]y\geq x[/itex]? If not, you should be integrating from [itex]0[/itex] to [itex]1-x[/itex] instead.
 
  • #3
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I know that y>=x-1
and if I do it for 0 to 1 I don't get in the end that the probabilty =1
 
  • #4
vela
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You were given [itex]x+y \le 1[/itex], so, moving the x over to the other side, you get [itex]y \le 1-x[/itex], right?
 
  • #5
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right... so that's why I chose the integral limits for dy to be from x to 1-x.
Is it wrong?
 
  • #6
gabbagabbahey
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right... so that's why I chose the integral limits for dy to be from x to 1-x.
Is it wrong?
Again, why are you using [itex]x[/itex] as your lower limit? You are told that [itex]y\geq0[/itex], not [itex]y\geq x[/itex]
 
  • #7
vela
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I know that y>=x-1
and if I do it for 0 to 1 I don't get in the end that the probabilty =1
right... so that's why I chose the integral limits for dy to be from x to 1-x.
Is it wrong?
Your response to gabbagabbahey's question was confusing. In your previous post, you answered, it was because you knew [itex]y \ge x-1[/itex]. The "x-1" suggested to me you somehow got that condition from the constraint [itex]x+y \le 1[/itex], and my point was that the constraint can only lead to [itex]y \le 1-x[/itex], the upper limit of the integral. There's no way you can get [itex]y \ge x-1[/itex] from it.

Even if you did have [itex]y \ge x-1[/itex], your answer didn't make sense. Why would the lower limit of the integral be x? Shouldn't it be x-1 if your inequality were true?
 

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