Statistics expectation problem involving circle.

[peripatein]

Hi,

Homework Statement

A circle of radius r is as shown in the attached diagram. I am asked to first express X as a function of θ, then to compute E(X). It is also stated that θ obeys U[0,2π].

Homework Equations

The Attempt at a Solution

Through simple trigonometry I have found X = 2rsin(θ/2). As θ distributes evenly, I know that the density of θ should be 1/2π. What is the density of x then? Am I supposed to simply substitute 2arcsin(x/2r)? Or should it be approached by first computing E(θ) and trying to hence infer E(X)?
I'd appreciate an advice.

 

[tiny-tim]

hi peripatein!

look at the definition …

if X(θ) is a function of θ, and if the distribution of θ is f(θ),

then what is the definition of E(X) ?

 

[peripatein]

The definition of E(X) is integral [would the boundaries be 1 to 2r?] x*f(x) dx, where f(x) is the probability density of X. Is this correct?

 

[tiny-tim]

The definition of E(X) is integral [would the boundaries be 1 to 2r?] x*f(x) dx, where f(x) is the probability density of X. Is this correct?

yees, but it's f(θ) that you know, not f(x) …

so what is E(X) expressed as an integral over θ ? ​

 

[peripatein]

E(X) = ∫(2rsin(θ/2)/2π)(rcos(θ/2))dθ from 0 to 2π?

 

[tiny-tim]

no

stick to the definition for the moment …

what is the definition of E(X) if X = X(θ), and the distribution of θ is f(θ) ?

 

[peripatein]

I am not sure, tiny-tim :-(.
Should I look for the inverse function of X(θ)?
I mean, the definition reads E[g(X)] = integral (from -inf. to +inf.) g(x)f(x)dx. I understand that as E[θ(X)] = θ(X)*f(x)dx. I am probably mistaken, yet I am not sure how to approach it.
Here's another attempt, which makes not much sense to me but for the sake of trying: might it be E(X) = int X(θ)*f(θ)dθ?

 

[tiny-tim]

IHere's another attempt, which makes not much sense to me but for the sake of trying: might it be E(X) = int X(θ)*f(θ)dθ?

yes, that's correct …

E(X) = ∫ X(θ) f(θ) dθ

… this is exactly the same as the definition you mentioned earlier, ∫ x f(x) dx, but using the distribution of θ rather than the distribution of x (remember, the two f's are different!)

so in this case, E(X) = ∫ 2rsin(θ/2) f(θ) dθ,

which is … ? ​

 

[peripatein]

4r/pi?
How may I now determine whether X obeys an even distribution U?

 

[tiny-tim]

show how you got it

 

[peripatein]

Well, shouldn't it be E(X) = ∫ 2rsin(θ/2) f(θ) dθ = ∫ (r/pi)sin(θ/2) dθ between 0 and 2pi?
How may I now determine whether X obeys an even distribution U?

 

[tiny-tim]

Well, shouldn't it be E(X) = ∫ 2rsin(θ/2) f(θ) dθ = ∫ (r/pi)sin(θ/2) dθ between 0 and 2pi?

correct

How may I now determine whether X obeys an even distribution U?

it obviously doesn't

θ is evenly distributed, so sinθ/2 isn't

(why are you asking? doesn't the question only ask for E(X) ? )

 

[peripatein]

Now, the second part of that question, which is not related to the first, asks for a proof that
∫ (from 0 to ∞) xⁿe^-x dx = n! for every n > 0, and the hint is to find a recursive argument for I_n and an initial condition.
I have found I_n = (I_{n + 1} + xⁿe^-x + nx^n-1e^-x) / (n(n-1)), where x changes from 0 to ∞, but am not sure how to proceed. Could you advise?

 

[tiny-tim]

Now, the second part of that question, which is not related to the first, asks for a proof that
∫ (from 0 to ∞) xⁿe^-x dx = n! for every n > 0, and the hint is to find a recursive argument for I_n and an initial condition.
I have found I_n = (I_{n + 1} + xⁿe^-x + nx^n-1e^-x) / (n(n-1)), where x changes from 0 to ∞, but am not sure how to proceed. Could you advise?

(how did you get that ?? )

anyway, you shouldn't have any x's in the equation

x is a dummy variable, the only "real" variable is n​

 

[peripatein]

Well, integration by parts, twice. Setting the original integral to be I_n, I let u = xⁿ and dv be e^-x. In the second integration by parts I let u = x^n-1 and dv = e^-x.
Does that make more sense now? Is that how it is to be handled? The boundaries should be between 0 and infinity, but I don't know how to incorporate that into the expression (I mean, what do you get for (xⁿe^-x + nx^n-1e^-x) between 0 and inf.? How do you evaluate that?

 

[tiny-tim]

Well, integration by parts, twice.

why not just once?

The boundaries should be between 0 and infinity, but I don't know how to incorporate that into the expression (I mean, what do you get for (xⁿe^-x + nx^n-1e^-x) between 0 and inf.? How do you evaluate that?

i] what is xⁿe^-x when x = 0 ?

ii] what is (the limit of) xⁿe^-x when x = ∞ ?

 

[peripatein]

You're right, silly me! A single integration by parts led me to the right answer: I_n+1 = (n+1)I_n.
Now, suppose I define f(x) = (xⁿe^-x)/n!. I have found E(X) = n + 1 and V(X) = 0.
Would you please confirm that?

 

[tiny-tim]

Now, suppose I define f(x) = (xⁿe^-x)/n!. I have found E(X) = n + 1

yes that looks correct

and V(X) = 0.

what is V(X) ?

 

[peripatein]

Variance (=E(x^2) - (E(x))^2)
Is it not zero in this case?

 

[tiny-tim]

no

how did you get E(x²) ?

 

[peripatein]

My bad, I made a mistake along the way. How's the following:
E(X²) = ∫ (from zero to infinity) (x²xⁿe^-x)/n!dx, right? And that is I_n+2/n!, which is in turn (n+2)(n+1), is it not? Now, (n+2)(n+1) - (n+1)² = n+1 = Var(X). Is it correct now?

 

[tiny-tim]

yes! :rofl:

you see now how important it is to write everything out carefully?

 

[peripatein]

Yes, tiny-tim, thank you very much! :-)