1. ## Just 2 questions...

Hello everyone
Would appreciate help with these two, especially the second one.

The joint densities of $S = X + Y$ and $T = \frac{X}{Y}$ is given by:

$f_{ST}(s,t) = \frac{\lambda ^2 s}{(1+t)^2} e^{- \lambda s}$

$s \geq 0 \ ; \ t \geq 0$

where $X$ and $Y$ are independent exponential random variables with parameter $\lambda$

Prove that S and T are independent by finding $f_{S}(s)$ and $f_{T}(t)$ and showing that $(f_{S}(s))(f_{T}(t)) = f_{ST}(s,t)$
Let $U_{1}$ and $U_{2}$ be independent and uniform on $[0,1]$. Find and sketch the density function of $S = U_{1} + U_{2}$
Once again I'd also like to ask if someone can suggest a book with sufficient examples and explanations regarding this kind of work. The book I have has very few examples, and even just as few explanations.

2. For the second the density of $S$ is the convolution of the densities of $U_1$ and $U_2$, which gives the triangular distribution on $[0,2]$ (symmetric about $1$)

RonL

3. The joint densities of $S = X + Y$ and $T = \frac{X}{Y}$ is given by:
$f_{ST}(s,t) = \frac{\lambda ^2 s}{(1+t)^2} e^{- \lambda s}$
$s \geq 0 \ ; \ t \geq 0$
where $X$ and $Y$ are independent exponential random variables with parameter $\lambda$

The marginals are:

$f_S(s)=\int_0^{\infty} f_{ST}(s,t) dt=s \lambda^2 e^{-\lambda s}$

and:

$f_T(t)=\int_0^{\infty} f_{ST}(s,t) ds=\frac{1}{(t+1)^2}$

and the rest follows (this is obvious as $f_{ST}$ seperates into a product of a function of $s$ and a function of $t$)

RonL

4. Originally Posted by CaptainBlack
The joint densities of $S = X + Y$ and $T = \frac{X}{Y}$ is given by:
$f_{ST}(s,t) = \frac{\lambda ^2 s}{(1+t)^2} e^{- \lambda s}$
$s \geq 0 \ ; \ t \geq 0$
where $X$ and $Y$ are independent exponential random variables with parameter $\lambda$

The marginals are:

$f_S(s)=\int_0^{\infty} f_{ST}(s,t) dt=s \lambda^2 e^{-\lambda s}$

and:

$f_T(t)=\int_0^{\infty} f_{ST}(s,t) ds=\frac{1}{(t+1)^2}$

and the rest follows (this is obvious as $f_{ST}$ seperates into a product of a function of $s$ and a function of $t$)

RonL
Shouldn't they both be negative?

5. Originally Posted by CaptainBlack
For the second the density of $S$ is the convolution of the densities of $U_1$ and $U_2$, which gives the triangular distribution on $[0,2]$ (symmetric about $1$)

RonL
Still a little bit lost on this one. The lecturer said we'll hear about convolutions a bit later, and not to worry about it yet... (I'll have to go look up what a triangular distribution is as well.)

Originally Posted by CaptainBlack
The joint densities of $S = X + Y$ and $T = \frac{X}{Y}$ is given by:
$f_{ST}(s,t) = \frac{\lambda ^2 s}{(1+t)^2} e^{- \lambda s}$
$s \geq 0 \ ; \ t \geq 0$
where $X$ and $Y$ are independent exponential random variables with parameter $\lambda$

The marginals are:

$f_S(s)=\int_0^{\infty} f_{ST}(s,t) dt=s \lambda^2 e^{-\lambda s}$

and:

$f_T(t)=\int_0^{\infty} f_{ST}(s,t) ds=\frac{1}{(t+1)^2}$

and the rest follows (this is obvious as $f_{ST}$ seperates into a product of a function of $s$ and a function of $t$)

RonL
This has helped a lot, thank you Captain.

Originally Posted by tukeywilliams
Shouldn't they both be negative?
Here's just a guess... they're exponential so they must be positive?

EDIT: Wait, for some bizarre reason I also think they must be both negative. I tried working it out by myself just now...

6. Originally Posted by tukeywilliams
Shouldn't they both be negative?
They are pdf's of the marginal distributions and so have to be positive (and integrate up to 1).

RonL

7. Originally Posted by janvdl
Still a little bit lost on this one. The lecturer said we'll hear about convolutions a bit later, and not to worry about it yet... (I'll have to go look up what a triangular distribution is as well.)
..
This is a bit fiddly to do as the convolution is peicewise continuous, but the convolution is:

$f_S(s)=\int_{-\infty}^{\infty}p_{U_1}(u_1)p_{U_2}(s-u_1) \ dx_1$

and using the properties of the uniform density:

$f_S(s)=\int_{-\infty}^{\infty}p_{U_1}(u_1)p_{U_2}(s-u_1) \ dx_1=\int_{0}^{1}p_{U}(s-u) \ du$

RonL

8. Originally Posted by CaptainBlack
The joint densities of $S = X + Y$ and $T = \frac{X}{Y}$ is given by:
$f_{ST}(s,t) = \frac{\lambda ^2 s}{(1+t)^2} e^{- \lambda s}$
$s \geq 0 \ ; \ t \geq 0$
where $X$ and $Y$ are independent exponential random variables with parameter $\lambda$

The marginals are:

$f_S(s)=\int_0^{\infty} f_{ST}(s,t) dt=s \lambda^2 e^{-\lambda s}$

and:

$f_T(t)=\int_0^{\infty} f_{ST}(s,t) ds=\frac{1}{(t+1)^2}$

and the rest follows (this is obvious as $f_{ST}$ seperates into a product of a function of $s$ and a function of $t$)

RonL
To Jan:

A reference you might find helpful:

http://www.dartmouth.edu/~chance/tea...k/Chapter7.pdf

Given your current knowledge, I think it’s instructive for you to think about how to calculate the pdf of $T = \frac{X}{Y}$.

The cdf of T is given by $F (t) = \Pr(T \leq t) = \Pr \left( \frac{X}{Y} \leq t \right) = \Pr(X \leq t Y)$ since Y > 0.

Therefore:

$F (t) = \int_{y = 0}^{\infty} \int_{x=0}^{x=ty} g(x,y) \, dx \, dy$

where $g(x, y)$ is the joint pdf of X and Y.

Since X and Y are independent random variables: $g(x,y) = \frac{1}{\lambda^2} e^{-x/ \lambda} \, e^{-y/ \lambda}$ for x > 0, y > 0, and is zero elsewhere.

Therefore:

$F (t) = \frac{1}{\lambda^2} \int_{y = 0}^{+ \infty} e^{-y/ \lambda} \int_{x=0}^{x=ty} e^{-x/ \lambda} \, dx \, dy$

$= \frac{1}{\lambda} \int_{y = 0}^{+ \infty} e^{-y/ \lambda} \left[ - e^{-x/ \lambda} \right]_{0}^{ty} \, dy$

$= \frac{1}{\lambda} \int_{y = 0}^{+ \infty} e^{-y/ \lambda} \left( 1 - e^{-ty/ \lambda} \right) \, dy$

$= \frac{1}{\lambda} \int_{y = 0}^{+ \infty} e^{-y/ \lambda} - e^{-y(1+t)/ \lambda} \, dy$

$= \left[ -e^{-y/ \lambda} + \frac{1}{1+t} \, e^{-y(1+t)/ \lambda} \right]_{0}^{+ \infty}$

$= 1 - \frac{1}{1 + t}$.

Therefore the pdf of T is given by $f(t) = \frac{dF}{dt} = \frac{1}{(1 + t)^2}$.

This is (of course) in agreement with the marginal distribution found using the given joint pdf of S and T:

$f(t) = \int_{0}^{+ \infty} \frac{\lambda^2 s}{(1 + t)^2}\, e^{- \lambda s} \, ds$ $= \frac{\lambda^2}{(1 + t)^2} \,$ $\int_{0}^{+ \infty}$ $s \, e^{- \lambda s} \, ds$ .....

After studying the definition of convolution and looking at CaptainB’s example you should attempt to get the pdf of S = X + Y.

9. Originally Posted by janvdl
Hello everyone
Would appreciate help with these two, especially the second one.

Quote:
Let and be independent and uniform on . Find and sketch the density function of

Once again I'd also like to ask if someone can suggest a book with sufficient examples and explanations regarding this kind of work. The book I have has very few examples, and even just as few explanations.
The second one is done as Example 7.3 in the link I provided. It confirms the answer given by CaptainB (not that confirmation was required lol!)

10. Originally Posted by CaptainBlack
For the second the density of $S$ is the convolution of the densities of $U_1$ and $U_2$, which gives the triangular distribution on $[0,2]$ (symmetric about $1$)

RonL
To Jan: It can also be done without ever knowing that you're essentially doing a convolution. To simplify the notation, I'll use S = U + V.

Since U and V are independent, their joint pdf is g(u, v) = (1)(1) = 1 for $0 \leq u \leq 1$, $0 \leq v \leq 1$, and zero elsewhere.

The cdf of S is given by $F(s) = \Pr(S \leq s) = \Pr(U + V \leq s) = \Pr(U \leq -V + s)$.

There are two cases to consider: $0 \leq s \leq 1$ and $1 \leq s \leq 2$.

Case 1: $0 \leq s \leq 1$.

$F(s) = \int_{v=0}^{s} \int_{u=0}^{-v+s} du \, dv = \int_{0}^{s} [u]_{0}^{-v+s} \, dv = \int_{0}^{s} -v + s \, dv = \left[ -\frac{1}{2} v^2 + sv\right]_{0}^{s} = \frac{1}{2} s^2$.

Therefore $f(s) = \frac{dF}{ds} = s$ for $0 \leq s \leq 1$.

Case 2: $1 \leq s \leq 2$.

$F(s) = \int_{v=0}^{s-1} \int_{u=0}^{1} du \, dv + \int_{v=s-1}^{1} \int_{u=0}^{u=-v+s} du \, dv = \, .....$

I'll leave the mopping up to you. You should get $f(s) = \frac{dF}{ds} = 2 - s$ for $1 \leq s \leq 2$.

And of course f(s) = 0 elsewhere. These are the same results as CaptainB got using the convolution theorem.

11. Originally Posted by mr fantastic
To Jan: It can also be done without ever knowing that you're essentially doing a convolution. To simplify the notation, I'll use S = U + V.

Since U and V are independent, their joint pdf is g(u, v) = (1)(1) = 1 for $0 \leq u \leq 1$, $0 \leq v \leq 1$, and zero elsewhere.

The cdf of S is given by $F(s) = \Pr(S \leq s) = \Pr(U + V \leq s) = \Pr(U \leq -V + s)$.

There are two cases to consider: $0 \leq s \leq 1$ and $1 \leq s \leq 2$.

Case 1: $0 \leq s \leq 1$.

$F(s) = \int_{v=0}^{s} \int_{u=0}^{-v+s} du \, dv = \int_{0}^{s} [u]_{0}^{-v+s} \, dv = \int_{0}^{s} -v + s \, dv = \left[ -\frac{1}{2} v^2 + sv\right]_{0}^{s} = \frac{1}{2} s^2$.

Therefore $f(s) = \frac{dF}{ds} = s$ for $0 \leq s \leq 1$.

Case 2: $1 \leq s \leq 2$.

$F(s) = \int_{v=0}^{s-1} \int_{u=0}^{1} du \, dv + \int_{v=s-1}^{1} \int_{u=0}^{u=-v+s} du \, dv = \, .....$

I'll leave the mopping up to you. You should get $f(s) = \frac{dF}{ds} = 2 - s$ for $1 \leq s \leq 2$.

And of course f(s) = 0 elsewhere. These are the same results as CaptainB got using the convolution theorem.
Thank you Mr F. I left this one out on the paper though. But I was the only one in class who did work.

12. Originally Posted by mr fantastic
To Jan:

A reference you might find helpful:

http://www.dartmouth.edu/~chance/tea...k/Chapter7.pdf

Given your current knowledge, I think it’s instructive for you to think about how to calculate the pdf of $T = \frac{X}{Y}$.

The cdf of T is given by $F (t) = \Pr(T \leq t) = \Pr \left( \frac{X}{Y} \leq t \right) = \Pr(X \leq t Y)$ since Y > 0.

Therefore:

$F (t) = \int_{y = 0}^{\infty} \int_{x=0}^{x=ty} g(x,y) \, dx \, dy$

where $g(x, y)$ is the joint pdf of X and Y.

Since X and Y are independent random variables: $g(x,y) = \frac{1}{\lambda^2} e^{-x/ \lambda} \, e^{-y/ \lambda}$ for x > 0, y > 0, and is zero elsewhere.
May I know how does one get the above g(x,y)?
I got $g(x,y) = {y^2 {\lambda}^2 \over (x + y)} e^{-x/ \lambda} \, e^{-y/ \lambda}$ after subst for s and t.

13. Originally Posted by chopet
May I know how does one get the above g(x,y)?
I got $g(x,y) = {y^2 {\lambda}^2 \over (x + y)} e^{-x/ \lambda} \, e^{-y/ \lambda}$ after subst for s and t.
...... and are independent exponential random variables with parameter ......
Therefore g(x, y) = f(x) f(y) = ......

14. Ok...think I missed a Jacobian term when converting from (s,t) -> (x,y)

$J = {{\partial s} \over {\partial x}} {{\partial t} \over {\partial y}} - {{\partial s} \over {\partial y}} {{\partial t} \over {\partial x}} = {{x+y} \over {y^2}}$

$
g(x,y) = {y^2 {\lambda}^2 \over (x + y)} e^{-x/ \lambda} \, e^{-y/ \lambda}*J = {\lambda}^2 e^{-x/ \lambda} \, e^{-y/ \lambda}$

still, the ${\lambda}^2$ term is wrong...

15. Originally Posted by chopet
Ok...think I missed a Jacobian term when converting from (s,t) -> (x,y)

$J = {{\partial s} \over {\partial x}} {{\partial t} \over {\partial y}} - {{\partial s} \over {\partial y}} {{\partial t} \over {\partial x}} = {{x+y} \over {y^2}}$

$
g(x,y) = {y^2 {\lambda}^2 \over (x + y)} e^{-x/ \lambda} \, e^{-y/ \lambda}*J = {\lambda}^2 e^{-x/ \lambda} \, e^{-y/ \lambda}$

still, the ${\lambda}^2$ term is wrong...
...... and are independent exponential random variables with parameter ......
Therefore g(x, y) = f(x) f(y) = ......
where $f(x) = \frac{1}{\lambda} \, e^{-x/\lambda}$ ....

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