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Probability- Exponential Distribution/Joint pdfs

  • Thread starter Zoe-b
  • Start date
98
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1. The problem statement, all variables and given/known data
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.


2. Relevant equations

I know how to find the the pdf of Y, Z seperately (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.


3. The attempt at a solution

So far I have:
FY,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
FY,Z(y,z) = 1 - exp(-y(λ+μ)) - exp(-λz) - exp(-μz) + exp(-μy-λz) + exp(-λy-μz)
for y<= z.

so
fY,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

Science Advisor
Homework Helper
Dearly Missed
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1. The problem statement, all variables and given/known data
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.


2. Relevant equations

I know how to find the the pdf of Y, Z seperately (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.


3. The attempt at a solution

So far I have:
FY,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
FY,Z(y,z) = 1 - exp(-y(λ+μ)) - exp(-λz) - exp(-μz) + exp(-μy-λz) + exp(-λy-μz)
for y<= z.

so
fY,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
 
98
0
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

Science Advisor
Homework Helper
Dearly Missed
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If X1 and X2 are independent, you can easily compute P{y < X1 < z & y < X2 < z}.

RGV
 
98
0
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

Science Advisor
Homework Helper
Dearly Missed
10,705
1,708
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
 
98
0
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) = FY,Z(y,z)

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

Ray Vickson

Science Advisor
Homework Helper
Dearly Missed
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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
 
98
0
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

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Homework Helper
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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

Science Advisor
Homework Helper
Dearly Missed
10,705
1,708
1. The problem statement, all variables and given/known data
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.


2. Relevant equations

I know how to find the the pdf of Y, Z seperately (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.


3. The attempt at a solution

So far I have:
FY,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
FY,Z(y,z) = 1 - exp(-y(λ+μ)) - exp(-λz) - exp(-μz) + exp(-μy-λz) + exp(-λy-μz)
for y<= z.

so
fY,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
 
98
0
(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 :)
 

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