# uniform random variables

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• Mar 7th 2009, 09:19 AM
cryptic26
uniform random variables
Let the variable be
Yc = theta_1 + theta_1*theta_2+ theta_1*theta_2 theta_3 + ....
where theta_i are i.i.d uniform random variables (i = 1,2,...) having

CDF such that F(x) = 0, x <=0. F(x) = x,
F(x) = 0<x<c and
F(x) = 1 , for x >=c.

Question,
1)for what values of c > 0 is E(Yc) finite.
2)Also, for what value of c > 0, Yc (sum converges with probability 1).
• Mar 7th 2009, 08:34 PM
meymathis
Ok so $\theta_i$ are IID, Uniform on [0,c]. And
$Y_c=\sum_{n=1}^\infty\prod_{m=1}^{n}\theta_m$

Well the expectation of the sum is the sum of the expectations. And the expectation of the product, if independent, is the product of the expectations. So

$E[Y_c] = E[\sum_{n=1}^\infty\prod_{m=1}^{n}\theta_m]$

$= \sum_{n=1}^\infty\prod_{m=1}^{n}E[\theta_m]$

$= \sum_{n=1}^\infty\prod_{m=1}^{n} c/2$

$= \sum_{n=1}^\infty (c/2)^n$

Which is a geometric series (minus the zeroth term). We know that converges for ...?

As for the second part, ask yourself this question. If the expectation is finite, could the probability of $Y_c$ being infinite be non-zero? So doesn't the answer to part 1 answer part 2?
• Mar 7th 2009, 10:15 PM
cryptic26
I did exactly as you have done now, but there is a small error in your E(X). It would be c^2/2 as f(x) = 1 for [0,c], which is somewhat different than the conventional f(x) of Uniform distribution which would be 1/c. However,the answer (I was told is wrong). Hence, I posted the question. Not sure what could be the catch?

Quote:

Originally Posted by meymathis
Ok so $\theta_i$ are IID, Uniform on [0,c]. And
$Y_c=\sum_{n=1}^\infty\prod_{m=1}^{n}\theta_m$

Well the expectation of the sum is the sum of the expectations. And the expectation of the product, if independent, is the product of the expectations. So

$E[Y_c] = E[\sum_{n=1}^\infty\prod_{m=1}^{n}\theta_m]$

$= \sum_{n=1}^\infty\prod_{m=1}^{n}E[\theta_m]$

$= \sum_{n=1}^\infty\prod_{m=1}^{n} c/2$

$= \sum_{n=1}^\infty (c/2)^n$

Which is a geometric series (minus the zeroth term). We know that converges for ...?

As for the second part, ask yourself this question. If the expectation is finite, could the probability of $Y_c$ being infinite be non-zero? So doesn't the answer to part 1 answer part 2?

• Mar 7th 2009, 10:35 PM
meymathis
Quote:

Originally Posted by cryptic26
I did exactly as you have done now, but there is a small error in your E(X). It would be c^2/2 as f(x) = 1 for [0,c], which is somewhat different than the conventional f(x) of Uniform distribution which would be 1/c. However,the answer (I was told is wrong). Hence, I posted the question. Not sure what could be the catch?

You have:
CDF such that F(x) = 0, x <=0. F(x) = x,
F(x) = 0<x<c and
F(x) = 1 , for x >=c.

Which I assumed was (since you said uniform in the title, I didn't look more closely):
CDF such that F(x) = 0 for x <=0.
F(x) = x, for 0<x<c and
F(x) = 1 , for x >=c.

Now that I look at this, this is NOT a CDF, unless c<1. A CDF has to be non-decreasing, and it most be 0 as $x\rightarrow -\infty$ and must be 1 and $x\rightarrow \infty$. If c>1 then F would not be non-decreasing.

You have uniform distribution in your title. Were you told that this was a problem concerning uniform distributions somewhere? This is certainly NOT a uniform distribution. If this is supposed to be about the uniform distributions, then I would think that the problem should have been stated as:
CDF such that F(x) = 0 for x <=0.
F(x) = x/c, for 0<x<c and
F(x) = 1 , for x >=c.
• Mar 7th 2009, 10:39 PM
meymathis
Quote:

Originally Posted by meymathis
Now that I look at this, this is NOT a CDF, unless c<1. A CDF has to be non-decreasing, and it most be 0 as $x\rightarrow -\infty$ and must be 1 as $x\rightarrow \infty$.

I made a typo which I underlined in italics above.
• Mar 8th 2009, 08:40 AM
cryptic26
If that is what you think, then write a letter to the author of the book who wrote this question :) . No one told you that it is a standard uniform variable. You just made that assumption.

Quote:

Originally Posted by meymathis
You have:

You have uniform distribution in your title. Were you told that this was a problem concerning uniform distributions somewhere? This is certainly NOT a uniform distribution. If this is supposed to be about the uniform distributions, then I would think that the problem should have been stated as:
CDF such that F(x) = 0 for x <=0.
F(x) = x/c, for 0<x<c and
F(x) = 1 , for x >=c.

• Mar 8th 2009, 08:45 AM
cryptic26
Sum and product of uniform variables.
I revised the title, in case it helps. This is what I wrote in my first post.

Let the variable be
Yc = theta_1 + theta_1*theta_2+ theta_1*theta_2 theta_3 + ....
where theta_i are i.i.d uniform random variables (i = 1,2,...) having the CDF that I mentioned in three cases.

This is where the uniform random variables comes into the title.

Quote:

Originally Posted by cryptic26
I think you are making an unnecessary fuss :) over the title. I wrote the question as it was in the text book, where it says we have uniform i.i.d random variables having CDF given as (three cases). If what I wrote is not clear, then the original question is no better.

Having said that, for c in (-inf, inf), the CDF is non decreasing.
As c >= inf , F(x) =1.
c <= -inf, F(x) = 0
and anywhere in between, F(x) = c, which is non decreasing. So, there is nothing wrong with the CDF function.

• Mar 8th 2009, 10:51 AM
cryptic26
Send to the author
I think you are making an unnecessary fuss :) over the title. I wrote the question as it was in the text book, where it says we have uniform i.i.d random variables having CDF given as (three cases). If what I wrote is not clear, then the original question is no better.

Having said that, for c in (-inf, inf), the CDF is non decreasing.
As c >= inf , F(x) =1.
c <= -inf, F(x) = 0
and anywhere in between, F(x) = c, which is non decreasing. So, there is nothing wrong with the CDF function.

Quote:

Originally Posted by meymathis
You have:
CDF such that F(x) = 0, x <=0. F(x) = x,
F(x) = 0<x<c and
F(x) = 1 , for x >=c.

Which I assumed was (since you said uniform in the title, I didn't look more closely):
CDF such that F(x) = 0 for x <=0.
F(x) = x, for 0<x<c and
F(x) = 1 , for x >=c.

Now that I look at this, this is NOT a CDF, unless c<1. A CDF has to be non-decreasing, and it most be 0 as $x\rightarrow -\infty$ and must be 1 and $x\rightarrow \infty$. If c>1 then F would not be non-decreasing.

You have uniform distribution in your title. Were you told that this was a problem concerning uniform distributions somewhere? This is certainly NOT a uniform distribution. If this is supposed to be about the uniform distributions, then I would think that the problem should have been stated as:
CDF such that F(x) = 0 for x <=0.
F(x) = x/c, for 0<x<c and
F(x) = 1 , for x >=c.

• Mar 8th 2009, 11:15 AM
cryptic26
Quote:

Originally Posted by meymathis
I made a typo which I underlined in italics above.

In fact, I think the part of the problem, is also to find the "c" for which the CDF is defined. So, you are saying that if c >1, then the given CDF is not monotonically increasing. That makes sense. This would mean both E(Yc) and Yc shall be finite.
• Mar 8th 2009, 02:43 PM
meymathis
Quote:

Originally Posted by cryptic26
Let the variable be
Yc = theta_1 + theta_1*theta_2+ theta_1*theta_2 theta_3 + ....
where theta_i are i.i.d uniform random variables (i = 1,2,...) having

CDF such that F(x) = 0, x <=0. F(x) = x,
F(x) = 0<x<c and
F(x) = 1 , for x >=c.

Question,
1)for what values of c > 0 is E(Yc) finite.
2)Also, for what value of c > 0, Yc (sum converges with probability 1).

You seem to be misunderstanding me on many accounts.

The problem as stated said "where theta_i are i.i.d uniform random variables". The phrase "Uniform random variables" has a very specific meaning in probability theory. It means they have the Uniform_distribution. It doesn't mean that they are the same. IID means independent and identically distributed. Why would someone need to say the $\theta_i$ are the same twice?

I never assumed they were standard uniform. What I said, and now I think you understand, is that if C>1, then we don't get valid distributions. The distribution of $\theta_i$ is only uniform (of any kind) if C=1. Otherwise it is NOT a uniform distribution. For 0<C<1 you would get valid CDF, but it would not be a uniform distribution. That is why, I think the author just forgot to put the "/c" in the CDF. Then you get a problem about uniform random variables, which was specifically stated in the text of the problem.

I think the problem in the book has a typo in it. What is so scandalous about that? I have never seen a book that did not have a typo in it.

You are confusing me on what the book actually says:
In one place you have (CDF #1):
Quote:

CDF such that F(x) = 0, x <=0. F(x) = x,
F(x) = 0<x<c and
F(x) = 1 , for x >=c.
and in another (CDF #2)
Quote:

As c >= inf , F(x) =1.
c <= -inf, F(x) = 0
and anywhere in between, F(x) = c, which is non decreasing. So, there is nothing wrong with the CDF function.
The first one doesn't make sense because you (the author?) say that F(x)=x, F(x)=0<x<c. Do you mean F(x)=x for 0<x<c?

The second one IS a CDF of a non-uniform RV. It is an RV with mass only at 0 and c. Now F is constant rather than linear in (0,c).

In another post you said
Quote:

I did exactly as you have done now, but there is a small error in your E(X). It would be c^2/2 as f(x) = 1 for [0,c], which is somewhat different than the conventional f(x) of Uniform distribution which would be 1/c. However,the...
Here you say the PDF is 1 in [0,C]. Which corroborates the amended CDF #1.

So if you don't believe that the Author made a mistake and this (amended CDF #1) is what they meant, fine. Then you did not calculate the expected value of this partially continuous, partially discrete distribution correctly. There is a certain amount of mass at C, unlike every other point in the distribution; $P(\theta_i=C)=1-C$ (whereas for all continuous RV's, this would be 0). This is easy to see since $F(C)=1$ but $F(C-\epsilon)=C-\epsilon$ So $P(\theta_i=C)=\lim_{\epsilon \rightarrow 0^+}F(C)-F(C-\epsilon)=1-C$.

So $E[\theta_i] = C^2/2+(1-C)C = C-C^2/2$
The first term you had right. But you were missing the discrete mass at C. To prove convergence, since C<=1, $|C-C^2/2|. Thus you can bound the infinite series with a geometric series. For C<1, that geometric series converges, and therefore so does the one you care about. If C=1, then $E[\theta_i] = 1/2$. You get a geometric series which converges.
• Mar 8th 2009, 02:47 PM
meymathis
Quote:

Originally Posted by cryptic26
In fact, I think the part of the problem, is also to find the "c" for which the CDF is defined. So, you are saying that if c >1, then the given CDF is not monotonically increasing. That makes sense. This would mean both E(Yc) and Yc shall be finite.

This confuses me too. If it is not a valid CDF, then "E(Yc) and Yc shall be finite" is nonsensical since $\theta_i$ are not random variables.
• Mar 8th 2009, 03:22 PM
cryptic26
Let me thank you for detailed response, first of all. Secondly, no, I am not the author. And thirdly, I wish I knew how to edit the posts instead of creating new posts. The confusion is partly on account of multiple posts on my account.

It is true that the problem could have a typo. But, at the same time, the problem could be intentionally the way it is to make confuse the students.

Here is what is in the book about the CDF of the theta(s), which are i.i.d uniform random variables. This is as stated in the book.

CDF such that F(x) = 0, x <=0.

F(x) = x, 0<x<c and
F(x) = 1 , for x >=c.

You correctly pointed out that F(x) is not monotonous for c > 1. Hence, F(x) to be CDF one has a boundary on c <=1.

The next part is the E(Yc) = Integral (x.dx) for limits from 0 to C. Hence, I got infinite series with c^2/2 + .... + ...
which is convergent as long as 0 < c <=1.

Also, if c <1, then the infinite series, Yc shall be convergent with probability 1. (which is the second part). Please correct me again, if I am wrong or something is unclear
• Mar 8th 2009, 03:25 PM
cryptic26
Quote:

Originally Posted by meymathis
This confuses me too. If it is not a valid CDF, then "E(Yc) and Yc shall be finite" is nonsensical since $\theta_i$ are not random variables.

Discarding something as nonsense without careful consideration can be a quite dangerous.
• Mar 8th 2009, 04:18 PM
meymathis
Your welcome. Let's try to step back. Here are the major points of the proof:

Ok so $\theta_i$ are IID, with distribution given by:
CDF such that F(x) = 0, x <=0.

F(x) = x, 0<x<c and
F(x) = 1 , for x >=c.

THIS IS NOT A CDF IF C>1 since F(x) would not be a non-decreasing function. An RV must be described by a valid CDF. Therefore this problem only makes sense if $C\leq 1$ since $Y_c$ IS UNDEFINED otherwise (it is defined as a function of ILL-DEFINED components, $\theta_i$. (It is nonsense to discuss $Y_c$ for $c>1$, which I say after careful consideration).

So:

$Y_c=\sum_{n=1}^\infty\prod_{m=1}^{n}\theta_m$

Well the expectation of the sum is the sum of the expectations. And the expectation of the product, if independent, is the product of the expectations.

In a previous post, I showed (or tried to show) that $E[\theta_i]=c-c^2/2$. It is not $c^2/2$ because this ignores all of the mass at c. $E[\theta_i]$ has a strangeish distribution since the CDF is given by a discontinuous function (for $C\neq 1$. This is a very important point that you need to think about. For always continuous functions like we normally work with $P(X=x_0)=0$ since $P(X=x_0) = \lim_{\epsilon \rightarrow o+} P(X\leq x_0)-P(X\leq x_0-\epsilon)=\lim_{\epsilon \rightarrow o+} F_X(x_0)-F_X(x_0-\epsilon)$. The last expression is 0 if F is continuous. But our F is not continuous at c (if $C\neq 1$). To compute $E[\theta_i]$ you have to mix your approach of continuous and discrete techniques. $E[\theta_i]=\int_0^c xdx+c\,P(\theta_i=c)=c^2/2 + c(1-c)$

So

$E[Y_c] = E[\sum_{n=1}^\infty\prod_{m=1}^{n}\theta_m]$

$= \sum_{n=1}^\infty\prod_{m=1}^{n}E[\theta_m]$

$= \sum_{n=1}^\infty\prod_{m=1}^{n} (c-c^2/2)$

$= \sum_{n=1}^\infty(c-c^2/2)^n$

If $0\leq c\leq 1$ then the fact that $c-c/2 \leq c-c^2/2 \leq c$ implies that $0\leq c-c^2/2 \leq 1$

Also, I didn't mean to imply you were the author. I meant to say either you or the author was saying....
• Mar 8th 2009, 04:55 PM
cryptic26
Makes sense totally. I was already thinking of the discontinuity problem of F(x). If C > 1, then although, it is still continuous at 1 but if C > 1 , then for all values of x > 1, F(x+h) = 1 where as F(x) = x.

In fact, let me tell you that this problem is from a manuscript that shall be hopefully a published text book. The author is a retired professor of Statistics. I am helping solving some problems. I also try and post some other problems that I am learning for my own edification. In fact, the author told me that this problem is not obvious as it seems and you need careful consideration to be able to understand. As such, it is a sub-part of another huge problem. Thanks.

Quote:

Originally Posted by meymathis
Your welcome. Let's try to step back. Here are the major points of the proof:

Ok so $\theta_i$ are IID, with distribution given by:
CDF such that F(x) = 0, x <=0.

F(x) = x, 0<x<c and
F(x) = 1 , for x >=c.

THIS IS NOT A CDF IF C>1 since F(x) would not be a non-decreasing function. An RV must be described by a valid CDF. Therefore this problem only makes sense if $C\leq 1$ since $Y_c$ IS UNDEFINED otherwise (it is defined as a function of ILL-DEFINED components, $\theta_i$. (It is nonsense to discuss $Y_c$ for $c>1$, which I say after careful consideration).

So:

$Y_c=\sum_{n=1}^\infty\prod_{m=1}^{n}\theta_m$

Well the expectation of the sum is the sum of the expectations. And the expectation of the product, if independent, is the product of the expectations.

In a previous post, I showed (or tried to show) that $E[\theta_i]=c-c^2/2$. It is not $c^2/2$ because this ignores all of the mass at c. $E[\theta_i]$ has a strangeish distribution since the CDF is given by a discontinuous function (for $C\neq 1$. This is a very important point that you need to think about. For always continuous functions like we normally work with $P(X=x_0)=0$ since $P(X=x_0) = \lim_{\epsilon \rightarrow o+} P(X\leq x_0)-P(X\leq x_0-\epsilon)=\lim_{\epsilon \rightarrow o+} F_X(x_0)-F_X(x_0-\epsilon)$. The last expression is 0 if F is continuous. But our F is not continuous at c (if $C\neq 1$). To compute $E[\theta_i]$ you have to mix your approach of continuous and discrete techniques. $E[\theta_i]=\int_0^c xdx+c\,P(\theta_i=c)=c^2/2 + c(1-c)$

So

$E[Y_c] = E[\sum_{n=1}^\infty\prod_{m=1}^{n}\theta_m]$

$= \sum_{n=1}^\infty\prod_{m=1}^{n}E[\theta_m]$

$= \sum_{n=1}^\infty\prod_{m=1}^{n} (c-c^2/2)$

$= \sum_{n=1}^\infty(c-c^2/2)^n$

If $0\leq c\leq 1$ then the fact that $c-c/2 \leq c-c^2/2 \leq c$ implies that $0\leq c-c^2/2 \leq 1$

Also, I didn't mean to imply you were the author. I meant to say either you or the author was saying....

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