Probability- Exponential Distribution/Joint pdfs

[Zoe-b]

Homework Statement

Let X₁, X₂ be exponential RVs with parameter λ, μ respectively. (The question does NOT say that they are independent but I think this must be a typo?! If not I have even less idea how to do the question).

Let Y= min {X₁, X₂}
and Z = max {X₁, X₂}

Let W = Z-Y
Calculate the joint pdfs of (Y,Z), (Z,W), (Y,W). Which pairs are independent.

Homework Equations

I know how to find the the pdf of Y, Z separately (via the cdf) but this doesn't seem to be directly relevant. Clearly Y,Z are not independent, so I think I need to find their joint pdf by first finding the joint cdf and then integrating.

The Attempt at a Solution

So far I have:
F_Y,Z(y,z) = P( Y < y, Z < z)
= P(Y,Z<y) + P(Y<y, y<Z<z)
= P(X₁<y, X₂<y) + P(X₁ <y, y<X₂<z) + P(X₂<y, y<X₁<z)

I really would like to know if this line is valid.. from here I used independence of Y,Z to find that
F_Y,Z(y,z) = 1 - exp(-y(λ+μ)) - exp(-λz) - exp(-μz) + exp(-μy-λz) + exp(-λy-μz)
for y<= z.

so
f_Y,Z(y,z) = 0 for y<z
= λμ(exp(-μy-λz) + exp(-λy-μz)) otherwise

This looks promising on the basis that the integral first wrt y from 0 to z, and then wrt z from 0 to infinity is 1.

But if I attempt to find the marginal distribution by integrating from 0 to ∞ wrt z, I get:
μexp(-μy) + λexp(-λy)
which is not what I would expect- I think the answer should be
(μ+λ)exp(-(μ+λ)y)

Am I even on the right lines at all? Its a much longer question but if this is wrong (as I fear it might be) there's not much point trying to plow through the rest using the wrong formulas..
Thank you!

 

[Ray Vickson]

Homework Statement

Let X₁, X₂ be exponential RVs with parameter λ, μ respectively. (The question does NOT say that they are independent but I think this must be a typo?! If not I have even less idea how to do the question).

Let Y= min {X₁, X₂}
and Z = max {X₁, X₂}

Let W = Z-Y
Calculate the joint pdfs of (Y,Z), (Z,W), (Y,W). Which pairs are independent.

Homework Equations

I know how to find the the pdf of Y, Z separately (via the cdf) but this doesn't seem to be directly relevant. Clearly Y,Z are not independent, so I think I need to find their joint pdf by first finding the joint cdf and then integrating.

The Attempt at a Solution

So far I have:
F_Y,Z(y,z) = P( Y < y, Z < z)
= P(Y,Z<y) + P(Y<y, y<Z<z)
= P(X₁<y, X₂<y) + P(X₁ <y, y<X₂<z) + P(X₂<y, y<X₁<z)

I really would like to know if this line is valid.. from here I used independence of Y,Z to find that
F_Y,Z(y,z) = 1 - exp(-y(λ+μ)) - exp(-λz) - exp(-μz) + exp(-μy-λz) + exp(-λy-μz)
for y<= z.

so
f_Y,Z(y,z) = 0 for y<z
= λμ(exp(-μy-λz) + exp(-λy-μz)) otherwise

This looks promising on the basis that the integral first wrt y from 0 to z, and then wrt z from 0 to infinity is 1.

But if I attempt to find the marginal distribution by integrating from 0 to ∞ wrt z, I get:
μexp(-μy) + λexp(-λy)
which is not what I would expect- I think the answer should be
(μ+λ)exp(-(μ+λ)y)

Am I even on the right lines at all? Its a much longer question but if this is wrong (as I fear it might be) there's not much point trying to plow through the rest using the wrong formulas..
Thank you!

I don't think your expressions for the joint distribution of (Y,Z) are correct. In fact, P{y < Y & Z < z} = P{y < X1 < z & y < X2 < z}.

RGV

 

[Zoe-b]

Ok.. well better to find out its wrong now as I said :P
Erm.. slightly confused by what you wrote, I think I agree that
P{y < Y & Z < z} = P{y < X1 < z & y < X2 < z}.

but seeing as Y,Z are not independent I then don't understand how I can relate this expression to P(Y<y, Z<z). Am I being incredibly slow?

 

[Ray Vickson]

If X1 and X2 are independent, you can easily compute P{y < X1 < z & y < X2 < z}.

RGV

 

[Zoe-b]

Yes I know that. My question was I don't understand the relationship between

P{y < Y & Z < z} and P(Y<y & Z<z)

Obviously P(y<Y) = 1 - P(Y<y) but the dependence of Y,Z means I can't use this..

 

[Ray Vickson]

Yes I know that. My question was I don't understand the relationship between

P{y < Y & Z < z} and P(Y<y & Z<z)

Obviously P(y<Y) = 1 - P(Y<y) but the dependence of Y,Z means I can't use this..

Your statement "Obviously P(y<Y) = 1 - P(Y<y)" is clearly wrong: we don't have A = 1-A for A = P(Y < y), as you are claiming.

I don't see your problem. If y = 3 and z = 5, are the numbers 3 and 5 dependent? Of course, Y and Z are dependent, but that is why I was careful to distinguish between y and Y and between z and Z.

RGV

 

[Zoe-b]

Your statement "Obviously P(y<Y) = 1 - P(Y<y)" is clearly wrong: we don't have A = 1-A for A = P(Y < y), as you are claiming.

I don't see your problem. If y = 3 and z = 5, are the numbers 3 and 5 dependent? Of course, Y and Z are dependent, but that is why I was careful to distinguish between y and Y and between z and Z.

RGV

Thats not what I wrote-
P(Y≥y) + P(Y<y) = 1
last time I checked... my notation made it confusing apologies- the capitals were the other way around.

However I'm definitely still missing something here.

P(y < Y & Z < z) = P(Y>y, Z<z) ≠ P(Y<y & Z<z) = F_Y,Z(y,z)

So knowing the thing on the left does not help me find the thing on the right.

 

[Ray Vickson]

Knowing P(Y>y & Z <z) is enough to determine the joint density f(y,z), but if I tell you how, I would be doing your homework.

RGV

 

[Zoe-b]

Its actually a tiny part of a very long question that I'm taking as part of an optional course. So you're hardly doing my homework. But anyway. Thanks.

 

[Ray Vickson]

Its actually a tiny part of a very long question that I'm taking as part of an optional course. So you're hardly doing my homework. But anyway. Thanks.

Sigh. I'll give it one more try. If you _did_ know the bivariate density f(u,v) of (Y,Z) at all relevant values of Y=u and Z=v, you could obtain P(Y>y & Z<x) as a two-dimensional integral of f. Write out this integral. Now it should be obvious how you can get f.

RGV

 

[Ray Vickson]

Homework Statement

Let X₁, X₂ be exponential RVs with parameter λ, μ respectively. (The question does NOT say that they are independent but I think this must be a typo?! If not I have even less idea how to do the question).

Let Y= min {X₁, X₂}
and Z = max {X₁, X₂}

Let W = Z-Y
Calculate the joint pdfs of (Y,Z), (Z,W), (Y,W). Which pairs are independent.

Homework Equations

I know how to find the the pdf of Y, Z separately (via the cdf) but this doesn't seem to be directly relevant. Clearly Y,Z are not independent, so I think I need to find their joint pdf by first finding the joint cdf and then integrating.

The Attempt at a Solution

So far I have:
F_Y,Z(y,z) = P( Y < y, Z < z)
= P(Y,Z<y) + P(Y<y, y<Z<z)
= P(X₁<y, X₂<y) + P(X₁ <y, y<X₂<z) + P(X₂<y, y<X₁<z)

I really would like to know if this line is valid.. from here I used independence of Y,Z to find that
F_Y,Z(y,z) = 1 - exp(-y(λ+μ)) - exp(-λz) - exp(-μz) + exp(-μy-λz) + exp(-λy-μz)
for y<= z.

so
f_Y,Z(y,z) = 0 for y<z
= λμ(exp(-μy-λz) + exp(-λy-μz)) otherwise

This looks promising on the basis that the integral first wrt y from 0 to z, and then wrt z from 0 to infinity is 1.

But if I attempt to find the marginal distribution by integrating from 0 to ∞ wrt z, I get:
μexp(-μy) + λexp(-λy)
which is not what I would expect- I think the answer should be
(μ+λ)exp(-(μ+λ)y)

Am I even on the right lines at all? Its a much longer question but if this is wrong (as I fear it might be) there's not much point trying to plow through the rest using the wrong formulas..
Thank you!

Sorry: I mis-read your expression for f(y,z) (recovering from a cold/flu and not at my best): your expression _is_ correct, but I thought the derivation more convoluted than need be. Much easier: {Y=y,Z=z} = {X1=y & X2 = z} or {X2=y & x1=z}. (Also, you would get this from the expression H(y,z) = P(Y>y,Z<z), using f(y,z) = -∂^2 H/ ∂y ∂z.)

Also: your marginal of Y was computed incorrectly: you should integrate over z from z = y to z = ∞.

RGV

 

[Zoe-b]

(Also, you would get this from the expression H(y,z) = P(Y>y,Z<z), using f(y,z) = -∂^2 H/ ∂y ∂z.)

Yes I had actually worked that out, but I was just about ready to give up when it gave me the same answer I got the first time. Thanks for all your help, I now have some chance of finishing the sheet in time :)