Statistics uniform distribution problem

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  • #1
CAH
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It's hard to type this out as there is a diagram and notation I can't find on the key board so I've attached an image of the question and answer. I've explained my solution below however I've also attached an image if it's too confusing with the lack of symbols!

Problem involves uniform distribution and integration.

I figured that E(feta) = (0+(pi/2))/2
And that cos(feta) ~ U[0, 1]
So E(4cosfeta) = 4 x 1/2
This is incorrect and If I integrate 4cos(feta) x 2/pi I get the right answer but I don't know why the previous method dosent work?

b(ii) I did P(X<= 3) = P(cos(feta) <= 0.75) and then = 0.75 (by my previous calculation of cos(feta) ~ U..)

Also why is p(cos(feta) < 0.75) = P(feta => 0.732..) why does the sign switch??
 

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  • #2
Ray Vickson
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It's hard to type this out as there is a diagram and notation I can't find on the key board so I've attached an image of the question and answer. I've explained my solution below however I've also attached an image if it's too confusing with the lack of symbols!

Problem involves uniform distribution and integration.

I figured that E(feta) = (0+(pi/2))/2
And that cos(feta) ~ U[0, 1]
So E(4cosfeta) = 4 x 1/2
This is incorrect and If I integrate 4cos(feta) x 2/pi I get the right answer but I don't know why the previous method dosent work?

b(ii) I did P(X<= 3) = P(cos(feta) <= 0.75) and then = 0.75 (by my previous calculation of cos(feta) ~ U..)

Also why is p(cos(feta) < 0.75) = P(feta => 0.732..) why does the sign switch??

You don't need to type much out, and you do not need to present the diagram. You just have a random variable of the form ##X = 4 \cos(\Theta)##, where ##\Theta## is uniformly distributed on ##(0, \pi/2)##. You want to compute ##EX## and ##P(X \leq 3)##.

In your simple approach to ##EX## you have made the fundamental error that beginners often make: for a NONLINEAR function ##h(\theta)## we have ##E h(\Theta) \neq h(E\Theta)##, usually. Your error was to assume that was a true equation, but most often it is not. In this particular case, it is definitely not true.

Think about a simple case of a discrete random variable ##Y##, taking values ##y_i## with probabilities ##p_i##; for example, ##P(Y=y_1) = P(Y = y_2) = 1/2##. For a function f(y) we have ##Ef(Y) = \sum p_i f(y_i)##, but ##f(EY) = f(\sum p_i y_i)##. Generally, you will not have ##\sum p_i f(y_i) = f(\sum p_i y_i)## unless ##f## is a linear function of the form ##f(y) = ay + b##.

As you said, ##P(X \leq 3) = P(\cos(\Theta) \leq 3/4)##. Now look at the graph of ##y = \cos(\theta)## on the interval ##0 \leq \theta \leq \pi/2##. What does the region ##y \leq 0.75## look like on the ##\theta##-axis?

You might wonder: why did I write ##\Theta## sometimes and ##\theta## at other times? It was not a typo: I was respecting the mostly-accepted standards of probability writing, whereby a random variable is typically denoted by an upper-case letter and its possible values by the corresponding lower case letter. So, the random variable ##\Theta## takes values ##\theta## that lie in the interval ##(0,\pi/2)##. Your book apparently did not use that convention, at least in this case.
 
  • #3
Stephen Tashi
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And that cos(feta) ~ U[0, 1]
As Ray said, cos(feta) is not a linear function of feta. So the fact that feta has a uniform distribution does not imply that cos(feta) has a uniform distribution.
 
  • #4
BvU
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Could you good folks indulge me and not write feta ( a Greek kind of cheese ) but theta ( a Greek letter th ). It 'urts me eyes! :rolleyes:

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