Joint density function of iid exponential distribution?

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flumefan1
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Let random variables X1, X2 and X3 be independent and identically distributed according to the exponential distribution with rate λ. Let Y1=X1, Y2=X1+X2, and Y3=X1+X2+X3.
(a) Find the joint density function of Y1, Y2 and Y3.
(b) Find the marginal density of Y3.
What do I do here? I don't really know where to start. Thank you for any help!
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RDKGames
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(Original post by flumefan1)
Let random variables X1, X2 and X3 be independent and identically distributed according to the exponential distribution with rate λ. Let Y1=X1, Y2=X1+X2, and Y3=X1+X2+X3.
(a) Find the joint density function of Y1, Y2 and Y3.
(b) Find the marginal density of Y3.
What do I do here? I don't really know where to start. Thank you for any help!
Joint density function here is just

P(Y1=y1 AND Y2=y2 AND Y3=y3)

Are Y1,Y2,Y3 independent? If so, this has a nice simplification.
Last edited by RDKGames; 4 days ago
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flumefan1
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(Original post by RDKGames)
Joint density function here is just

P(Y1=y1 AND Y2=y2 AND Y3=y3)

Are Y1,Y2,Y3 independent? If so, this has a nice simplification.
How did you get that?
I don't know I would assume that they are as X1 X2 and X3 are independent but I don't see how Y1 Y2 and Y3 exist? What's the point of them. Why can't we work with the X's?
Last edited by flumefan1; 4 days ago
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RDKGames
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(Original post by flumefan1)
How did you get that?
I don't know I would assume that they are as X1 X2 and X3 are independent but I don't see how Y1 Y2 and Y3 exist? What's the point of them. Why can't we work with the X's?
Ignore the above I forgot the exponential is a continuous r.v.

Just check the def you have. From Wikipedia it says the JOINT pdf is the derivative of the CDF with respect to each variable.

So that’s what you need to find first.

F(y1,y2,y3) = P(Y1<y1 , Y2<y2 , Y3<y3)
Last edited by RDKGames; 4 days ago
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flumefan1
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(Original post by RDKGames)
Ignore the above I forgot the exponential is a continuous r.v.

Just check the def you have. From Wikipedia it says the JOINT pdf is the derivative of the CDF with respect to each variable.

So that’s what you need to find first.

F(y1,y2,y3) = P(Y1<y1 , Y2<y2 , Y3<y3)
Ah okay thank you. How do I find those?
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RDKGames
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(Original post by flumefan1)
Ah okay thank you. How do I find those?
Have you not seen examples?

Consider a two variable case.

P(Y_1 &lt; y_1,Y_2&lt;y_2) = P(X_1&lt;y_1,X_1+X_2&lt;y_2)

If X1 takes on value x1 then the above is the same as

\displaystyle P(X_1&lt;y_1,X_2&lt;y_2-x_1) = \int_{x_1=0}^{y_1} \int_{x_2=0}^{y_2-x_1} f_{X_1X_2}(x_1,x_2) \ dx_1 dx_2

Exploit the fact that X1,X2 are iid and compute.
Last edited by RDKGames; 4 days ago
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flumefan1
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(Original post by RDKGames)
Have you not seen examples?

Consider a two variable case.

P(Y_1 &lt; y_1,Y_2&lt;y_2) = P(X_1&lt;y_1,X_1+X_2&lt;y_2)

If X1 takes on value x1 then the above is the same as

\displaystyle P(X_1&lt;y_1,X_2&lt;y_2-x_1) = \int_{x_1=0}^{y_1} \int_{x_2=0}^{y_2-x_1} f_{X_1X_2}(x_1,x_2) \ dx_1 dx_2

Exploit the fact that X1,X2 are iid and compute.
I probably have seen examples, but not understood them. Is it a triple integral? if so how do I solve triple integrals?
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RDKGames
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(Original post by flumefan1)
I probably have seen examples, but not understood them. Is it a triple integral? if so how do I solve triple integrals?
Yes it’s a triple integral. If you can do the double integral above then it’s just one extra integration step for the triple integral.

Can you compute double integrals?
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flumefan1
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(Original post by RDKGames)
Yes it’s a triple integral. If you can do the double integral above then it’s just one extra integration step for the triple integral.

Can you compute double integrals?
I think so, I don't really know what they are as I'm in first year and they are a second year concept but I know how to compute them e.g.

\displaystyle  \int_{0}^{1} \int_{0}^{1} k(x<font size="2">^2 + y)</font> \ dx dx =1 I can solve that to find k.
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RDKGames
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(Original post by flumefan1)
I think so, I don't really know what they are as I'm in first year and they are a second year concept but I know how to compute them e.g.

\displaystyle  \int_{0}^{1} \int_{0}^{1} k(x<font size="2">^2 + y)</font> \ dx dx =1 I can solve that to find k.
So your course doesn’t cover multiple integrals? Okay, a different approach is probably expected then.

What have you covered in your course?
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flumefan1
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(Original post by RDKGames)
So your course doesn’t cover multiple integrals? Okay, a different approach is probably expected then.

What have you covered in your course?
It covers multiple integrals. We briefly touched on them last semester but were told to ignore them as they'd come in second year. However they've come up in my first year stats module. I spoke with calculus lecturer and I know how to do them now sorry for the confusion!
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RDKGames
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(Original post by flumefan1)
It covers multiple integrals. We briefly touched on them last semester but were told to ignore them as they'd come in second year. However they've come up in my first year stats module. I spoke with calculus lecturer and I know how to do them now sorry for the confusion!
Okay, it would sound strange if it wouldn't cover multiple integrals. They are littered across the entire probability theory when you come to doing joint distributions.

Anyway, look at this two variable case. Since X_1,X_2 are iid it means that

f_{X_1,X_2}(x_1,x_2) = f_{X_1}(x_1) f_{X_2}(x_2) = \lambda e^{-\lambda x_1} \cdot \lambda e^{-\lambda x_2} = \lambda^2 e^{-\lambda(x_1+x_2)}

So the joint CDF is

\displaystyle F_{Y_1,Y_2}(y_1,y_2) = \int_{x_1 = 0}^{y_1} \int_{x_2 = 0}^{y_2 - x_1} \lambda^2 e^{-\lambda(x_1+x_2)} \ d x_1 d x_2

Integration in multiple variables works just like PARTIAL differentiation, just backwards of course.

You don't want to integrate with respect to x_1 first because it appears in one of the bounds.

Instead, integrate w.r.t x_2 and obtain

\displaystyle F_{Y_1,Y_2}(y_1,y_2) = \int_0^{y_1} \bigg[ -\lambda e^{-\lambda(x_1 + x_2)} \bigg]_{x_2 = 0}^{y_2 - x_1} \ dx_1 = \int_0^{y_1} -\lambda ( e^{-\lambda y_2} - e^{-\lambda x_1} ) \ dx_1

Then integrate the leftover stuff w.r.t x_1

\displaystyle F_{Y_1,Y_2}(y_1,y_2) = \left[ -\lambda x_1 e^{-\lambda y_2} - e^{-\lambda x_1} \bigg]_{x_1 = 0}^{y_1} = 1 - e^{-\lambda y_1} - \lambda y_1 e^{-\lambda y_2}

The joint PDF is then

f_{Y_1,Y_2}(y_1,y_2) = \dfrac{\partial^2 F_{Y_1,Y_2}}{\partial y_1 \partial y_2} = \lambda^2 e^{-\lambda y_2}
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