independent random variables and a uniform distribution...

**iamthemanyes** · July 31st 2011, 01:27 PM

If X, Y, and Z are independent and they follow a Uniform[0,T] distribution, then what is the probability the largest of the three is larger than the sum of the other two?

How can we compute the joint density of X+Y and X/Y?

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I have tried a variety of different things, but am getting answers with random variables in them (for the first part)... I'm not sure if this is okay or not. Thanks!

**chisigma** · August 1st 2011, 03:42 AM

Originally Posted by iamthemanyes

If X, Y, and Z are independent and they follow a Uniform[0,T] distribution, then what is the probability the largest of the three is larger than the sum of the other two?...

If X has p.d.f. $f_{X} (t)$ , Y has p.d.f $f_{Y} (t)$ and Z has p.d.f. $f_{Z} (t)$ then S=X+Y+Z has p.d.f. $f_{S}(t) = f_{X} (t)* f_{Y} (t)*f_{Z}(t)$ where '*' means convolution. If for semplicity sake we suppose T=1, then...

$\mathcal{L} \{f_{X} (t)\} = \mathcal{L} \{f_{Y} (t)\} = \frac{1-e^{-s}}{s}$ (1)

$\mathcal{L} \{f_{-Z} (t)\} = \frac{e^{s}-1}{s}$ (2)

... so that if S=X+Y-Z then...

$\mathcal{L} \{f_{S} (t)\} = \mathcal{L} \{f_{X} (t)\}\ \mathcal{L} \{f_{Y} (t)\}\ \mathcal{L} \{f_{-Z} (t)\} =$

$= \frac{(1-e^{-s})^{2}\ (e^{s}-1)}{s^{3}} = \frac{e^{s} -3 +3\ e^{-s} - e^{-2\ s}}{s^{3}}$ (3)

... so that is...

$f_{S} (t)= \frac{t^{2}}{2}\ \{\mathcad{u} (t+1) -3\ \mathcad{u} (t) + 3\ \mathcad{u} (t-1) - \mathcad{u} (t-2) \}$ (4)

... and the requested probability is...

$P\{S<0\} = \int_{-1}^{0} f_{S}(t)\ dt = \frac{1}{6}$ (5)

Kind regards

$\chi$ $\sigma$

**Moo** · August 1st 2011, 05:56 AM

Originally Posted by chisigma

If X has p.d.f. $f_{X} (t)$ , Y has p.d.f $f_{Y} (t)$ and Z has p.d.f. $f_{Z} (t)$ then S=X+Y+Z has p.d.f. $f_{S}(t) = f_{X} (t)* f_{Y} (t)*f_{Z}(t)$ where '*' means convolution. If for semplicity sake we suppose T=1, then...

$\mathcal{L} \{f_{X} (t)\} = \mathcal{L} \{f_{Y} (t)\} = \frac{1-e^{-s}}{s}$ (1)

$\mathcal{L} \{f_{-Z} (t)\} = \frac{e^{s}-1}{s}$ (2)

... so that if S=X+Y-Z then...

$\mathcal{L} \{f_{S} (t)\} = \mathcal{L} \{f_{X} (t)\}\ \mathcal{L} \{f_{Y} (t)\}\ \mathcal{L} \{f_{-Z} (t)\} =$

$= \frac{(1-e^{-s})^{2}\ (e^{s}-1)}{s^{3}} = \frac{e^{s} -3 +3\ e^{-s} - e^{-2\ s}}{s^{3}}$ (3)

... so that is...

$f_{S} (t)= \frac{t^{2}}{2}\ \{\mathcad{u} (t+1) -3\ \mathcad{u} (t) + 3\ \mathcad{u} (t-1) - \mathcad{u} (t-2) \}$ (4)

... and the requested probability is...

$P\{S<0\} = \int_{-1}^{0} f_{S}(t)\ dt = \frac{1}{6}$ (5)

Kind regards

$\chi$ $\sigma$

This is completely false. You can't consider that Z is the max of the 3, since there is randomness : the max can vary. A very simple way to notice that is that max(X,Y,Z) has a different distribution than X, Y or Z.
And who would use this symbol for Laplace transforms in probability ? We'd rather talk about mgf.
Calculations on inverse Laplace transforms are not common in probability anyway.

The problem would rather ask for $P(\max(X,Y,Z)>X+Y+Z-\max(X,Y,Z))$

By the way : you're giving a full "solution", but you're not explaining anything ! Isn't it the other way round usually ?

**chisigma** · August 3rd 2011, 01:48 AM

Effectively I didn't read carefully the question and what I computed is the probability $P\{Z>X+Y\}$ . It was requested the probability that the largest variable overcomes the sum of others two, i.e. the quantity...

$P\{Z>X+Y | Z= \text{max}\ (X,Y,Z)\} = \frac{P\{Z>X+Y\}}{P\{Z= \text{max}\ (X,Y,Z)\}} = \frac{\frac{1}{6}}{\frac{1}{3}} = \frac{1}{2}$ (1)

Regarding the application of the Laplace Tranform to this particular problem, in fact that only the application of the basic properties that a random variable...

$X= X_{1} + X_{2} + ...+ X_{n}$ (2)

... which is the sum of n independent variables $X_{i}\ ,\ i=1,2,...,n$ with p.d.f. $f_{i}(t)\ ,\ i=1,2,...,n$ , has p.d.f. ...

$f(t)= f_{1}(t)\ *\ f_{2}(t)\ *\ ... *\ f_{n}(t)$ (3)

... where $*$ means convolution. It is well known that the convolution is, when possible, much easier to be performed is s domain instead of in t domain. An example in which this type approach is very efficient is the problem proposed in...

http://www.mathhelpforum.com/math-he...re-185513.html

Kind regards

$\chi$ $\sigma$

**Moo** · August 3rd 2011, 06:55 AM

Originally Posted by chisigma

Effectively I didn't read carefully the question and what I computed is the probability $P\{Z>X+Y\}$ . It was requested the probability that the largest variable overcomes the sum of others two, i.e. the quantity...

$P\{Z>X+Y | Z= \text{max}\ (X,Y,Z)\} = \frac{P\{Z>X+Y\}}{P\{Z= \text{max}\ (X,Y,Z)\}} = \frac{\frac{1}{6}}{\frac{1}{3}} = \frac{1}{2}$ (1)

I still disagree with this. What makes you think that there is only Z ? There can be an event when the max is Z, and another event when the max is X. That makes your computation incomplete. That is only one third of the desired probability.

An explanation for this :

P(max is > to the other two) = P(max is > to the other two , X = max)+P(max is > to the other two , Y = max)+P(max is > to the other two , Z = max)=3P(Z>X+Y | Z=max(X,Y,Z))

You're not telling how you came up with the pdf of S. Great job with the inverse Laplace transforms, but as I said previously, this is not a common technique we're taught in probability courses.

And what the heck is "u" in your first post ?
Sincerely, try to talk in a probabilistic way, your approaches are rather confusing than helping. And let's not talk about the full "solutions" you're always giving.

Regarding the application of the Laplace Tranform to this particular problem, in fact that only the application of the basic properties that a random variable...

Use MGF's.

$X= X_{1} + X_{2} + ...+ X_{n}$ (2)

... which is the sum of n independent variables $X_{i}\ ,\ i=1,2,...,n$ with p.d.f. $f_{i}(t)\ ,\ i=1,2,...,n$ , has p.d.f. ...

$f(t)= f_{1}(t)\ *\ f_{2}(t)\ *\ ... *\ f_{n}(t)$ (3)

... where $*$ means convolution.

Do you feel like repeating this all over again and again ?

It is well known that the convolution is, when possible, much easier to be performed is s domain instead of in t domain.

What does it mean ? You make no effort to talk in terms of probability stuff, people are not familiar with "s domain" "t domain" !

Convolution is not often used in this kind of problem. I computed the pdf of X+Y with the convolution method and it's just an ugly result that one can barely use.

An example in which this type approach is very efficient is the problem proposed in...

http://www.mathhelpforum.com/math-he...re-185513.html

Certainly not.

**chisigma** · August 5th 2011, 04:42 AM

Originally Posted by chisigma

... regarding the application of the Laplace Tranform to this particular problem, in fact that only the application of the basic properties that a random variable...

$X= X_{1} + X_{2} + ...+ X_{n}$ (2)

... which is the sum of n independent variables $X_{i}\ ,\ i=1,2,...,n$ with p.d.f. $f_{i}(t)\ ,\ i=1,2,...,n$ , has p.d.f. ...

$f(t)= f_{1}(t)\ *\ f_{2}(t)\ *\ ... *\ f_{n}(t)$ (3)

... where $*$ means convolution. It is well known that the convolution is, when possible, much easier to be performed is s domain instead of in t domain. An example in which this type approach is very efficient is the problem proposed in...

http://www.mathhelpforum.com/math-he...re-185513.html

Here the text...

...suppose X_1, X_2, ... , X_100 are independent random variables with common mean "mu" and variance "sigma squared." Let X be their average. What is the probability that |X - "mu" | is greater than or equal to 0.25?...

Very well!... in general if n is the number of random variables, the 'average' is by definition...

$X= \frac{X_{1}+ X_{2} + ... + X_{n}}{n}$ (1)

... i.e. their sum devided by n. Now we suppose that the $X_{k}$ are all uniformely distributed between 0 and 1 [so that we examine a particular case, not the general case...] so that is $\mu=\frac{1}{2}$ and $\sigma^{2}= \frac{1}{3}$ . If...

$f_{1}(t) = 1\, \text{if}\ 0<t<1\ ;\ 0\ \text{otherwise}$ (2)

... then ...

$\mathcal{L}\{f_{1}(t)\} = \frac{1-e^{-s}}{s}$ (3)

... so that , neglecting the term n in (1) and indicating with $f_{n}(t)$ the p.d.f of the X, is...

$\mathcal{L}\{f_{n}(t)\} = \frac{(1-e^{-s})^{n}}{s^{n}}$ (4)

... and performing the Inverse Laplace Transform of (4) we derive...

$f_{n} (t)= \frac{1}{(n-1)!}\ \sum_{k=0}^{n} (-1)^{k}\ \binom{n}{k}\ \gamma_{k} (t)$ (5)

... where...

$\gamma_{k}(t)= (t-k)^{n-1}\ \mathcad{u} (t-k)$ (6)

Of course for 'large n' the computation of (5) requires a computer and something like fifteen years ago I composed a specific computer program. Using this program we found that for n=100 is...

$P\{|X-50|>25\} = 2\ \int_{0}^{25} f_{100}(t)\ dt \sim 1.79\ 10^{-19}$ (7)

Kind regards

$\chi$ $\sigma$

**Moo** · August 5th 2011, 05:00 AM

Originally Posted by chisigma

Here the text...

...suppose X_1, X_2, ... , X_100 are independent random variables with common mean "mu" and variance "sigma squared." Let X be their average. What is the probability that |X - "mu" | is greater than or equal to 0.25?...

Very well!... in general if n is the number of random variables, the 'average' is by definition...

$X= \frac{X_{1}+ X_{2} + ... + X_{n}}{n}$ (1)

... i.e. their sum devided by n. Now we suppose that the $X_{k}$ are all uniformely distributed between 0 and 1 [so that we examine a particular case, not the general case...] so that is $\mu=\frac{1}{2}$ and $\sigma^{2}= \frac{1}{3}$ . If...

$f_{1}(t) = 1\, \text{if}\ 0<t<1\ ;\ 0\ \text{otherwise}$ (2)

... then ...

$\mathcal{L}\{f_{1}(t)\} = \frac{1-e^{-s}}{s}$ (3)

... so that , neglecting the term n in (1) and indicating with $f_{n}(t)$ the p.d.f of the X, is...

$\mathcal{L}\{f_{n}(t)\} = \frac{(1-e^{-s})^{n}}{s^{n}}$ (4)

... and performing the Inverse Laplace Transform of (4) we derive...

$f_{n} (t)= \frac{1}{(n-1)!}\ \sum_{k=0}^{n} (-1)^{k}\ \binom{n}{k}\ \gamma_{k} (t)$ (5)

... where...

$\gamma_{k}(t)= (t-k)^{n-1}\ \mathcad{u} (t-k)$ (6)

Of course for 'large n' the computation of (5) requires a computer and something like fifteen years ago I composed a specific computer program. Using this program we found that for n=100 is...

$P\{|X-50|>25\} = 2\ \int_{0}^{25} f_{100}(t)\ dt \sim 1.79\ 10^{-19}$ (7)

Kind regards

$\chi$ $\sigma$

Calling a method efficient when another one can give you the answer in a few lines is absurd. Use the CLT if you know the distribution (or even the first two moments) and go back home with your Laplace transforms.

You're insisting in using a method that is long, useless and inappropriate.

Your posts with that method don't even mention how you get the density, no one can know if it's correct or not, unless having studied inverse Laplace transforms, which is certainly not a prerequisite for probability theory at this level.

**mr fantastic** · August 5th 2011, 05:13 AM

I think the OP now has sufficient information to answer his/her question. If not, s/he can pm me.

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