View Full Version : Conditional expectation question

busstud123

03-01-2010, 12:10 PM

Hey,

I would really appreciate your help with the following question:

X and Y each has a uniform distribution on [-u,u] and the two are independent.

could you please show how to compute E(X|X<Y)?

Thank you very much for your help!

Here's how far I got:

Pr(X=u|X<Y) = Pr(X=u,X<Y) / Pr(X<Y) = Pr(X=u,u<Y) / Pr(u<Y) = (since X and Y are independent) Pr(X=u)*Pr(Y>u) / Pr(Y>u) = Pr(X=u)

That doesn't seem to make intuitive sense to me though.

Martingale

03-01-2010, 07:28 PM

Hey,

I would really appreciate your help with the following question:

X and Y each has a uniform distribution on [-u,u] and the two are independent.

could you please show how to compute E(X|X<Y)?

Thank you very much for your help!

Here's how far I got:

Pr(X=u|X<Y) = Pr(X=u,X<Y) / Pr(X<Y) = Pr(X=u,u<Y) / Pr(u<Y) = (since X and Y are independent) Pr(X=u)*Pr(Y>u) / Pr(Y>u) = Pr(X=u)

That doesn't seem to make intuitive sense to me though.

Are X,Y continuous uniform or discrete uniform?

I guess it should be continuous uniform on [-u, u]

Note ∀y∈(-u, u], ∀x∈[-u, y),

Pr{X ≤ x|X < y} = Pr{X ≤ x, X < y}/Pr{X < y} = Pr{X ≤ x}/Pr{X < y}

= [(x + u)/2u]/[(y + u)/2u] = (x + u)/(y + u)

fX|X<y(x) = dPr{X ≤ x|X < y}/dx = 1/(y + u)

*A simpler intuitive way is to note that X|X < y is just another uniform random variable

on [-u, y)

E[X|X < y] = ∫xdx/(y + u) = [y^2 - (-u)^2]/[2(y + u)] = (y - u)/2

As a result, y∈(-u, u]

E[X|X < Y] = ∫E[X|X < Y, Y = y]fY(y)dy = ∫[(y - u)/2][1/2u]dy

= (1/4u)[u^2/2 - u*u - (-u)^2/2 + u(-u) = -u/2

busstud123

03-02-2010, 02:38 PM

Thank you very much. That's great! I see the approach now.

Martingale

03-02-2010, 02:39 PM

I guess it should be continuous uniform on [-u, u]

Note ∀y∈(-u, u], ∀x∈[-u, y),

Pr{X ≤ x|X < y} = Pr{X ≤ x, X < y}/Pr{X < y} = Pr{X ≤ x}/Pr{X < y}

= [(x + u)/2u]/[(y + u)/2u] = (x + u)/(y + u)

fX|X<y(x) = dPr{X ≤ x|X < y}/dx = 1/(y + u)

*A simpler intuitive way is to note that X|X < y is just another uniform random variable

on [-u, y)

E[X|X < y] = ∫xdx/(y + u) = [y^2 - (-u)^2]/[2(y + u)] = (y - u)/2

As a result, y∈(-u, u]

E[X|X < Y] = ∫E[X|X < Y, Y = y]fY(y)dy = ∫[(y - u)/2][1/2u]dy

= (1/4u)[u^2/2 - u*u - (-u)^2/2 + u(-u) = -u/2

Hmmm.... I keep getting -u/3

busstud123

03-02-2010, 02:52 PM

Would it be ok for you to show your steps?

Dragan

03-03-2010, 01:22 AM

Hmmm.... I keep getting -u/3

Well, I also happen to keep getting -u/3. :)

busstud123

03-03-2010, 12:22 PM

Thanks for posting guys. Is there any chance you could make your derivation publicly available? Or email me privately?

Dragan

03-06-2010, 02:46 AM

Thanks for posting guys. Is there any chance you

could make your derivation publicly available?

Look here:

Let: U = min(X,Y), where min(X,Y)<=z.

Pr{min(X,Y) > z} = Pr{X>z} Intersect Pr{Y>z}.

Pr{U>z} = Pr{X>z}*Pr{Y>z}.

Pr{U>=z} = (1 - Fx(z))*(1 - Fy(z)).

Fu(z) = 1 - (1 - Fx(z))*(1 - Fy(z)).

Thus,

F_min(x,y) = Fx(z) + Fy(z) - Fx(z)*Fy(z).

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