[ns] Poisson process
Saikat Ray
raysaikat@lycos.com
Thu Jan 1 18:35:01 2004
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M(t10-t0) will be a Poisson variable with rate 10*lambda
(similarly M(t20-t0)). However, M(t10-t0)/10 is clearly not a
Poisson variable since the ratio may not even be an integer
(recall that a Poisson variable must take only integer values).
>From elementary calculations, Var(X1)=lambda/10 and
Var(X2)=lambda/20.
--------- Original Message ---------
DATE: Thu, 1 Jan 2004 13:33:22
From: Jing Zhang <jingz002000@yahoo.com>
To: raysaikat@lycos.com
Cc: ns-users@ISI.EDU
I have a new case different from what we discussed before.
Suppose I have a time sequence t0, t1, t2, ..., ti, tj, ...tn, ... and tj - ti = 1 second. The traffic arrival process is a Poisson process along the whole time axis. I define
X1(t) = M(t10-t0)/10 and X2(t) = M(t20-t0)/20, where M(t10-t0) means the number of arrivals during (t10-t0) period. Then according to my simulation, the variance of X1(t) and X2(t) are different. And both X1(t) and X2(t) are not Poisson processes. Could anybody tell me why X1(t) and X2(t) are not just the same processes as all the individual processes during (tj-ti)?
I appreciate your help.
Jing
Saikat Ray <raysaikat@lycos.com> wrote:
If you observe a random process for a time period 'T', and the number of arrivals during this period happen to have the same mean and variance, you CANNOT conclude that the process is a Poisson Process. There are many many processes whose mean and variances are independent and they may "happen to be" equal.
The simplest way to ascertain whether a given arrival process is Poisson, is to take a look at its inter-arrival times. If the distribution of the inter-arrival times are exponential (for simulations, you may choose to plot the histogram for a large number of samples, say 10000), as well as they are independent (this part is hard to verify by measurement), then the process is Poisson. But, certainly the first test would be to plot the histogram of the interarrival times and compare with the theoretical pdf.
I do not know whether your current mail is related to your last mail; but I think I should clarify one more thing. As I pointed out last time, if you simply average the cumulative number of arrivals of two independent Poisson processes, the result is NOT a Poisson process. However, if X(t) and Y(t) are two independent Poisson Processes and X1(t) is created from X(t) by sending an arrival at X to X1 with probability 'p1', and similarly for Y1(t); both X1(t) and Y1(t) are Poisson processes on their own right and since X(t) and Y(t) were independent, so are
X1(t) and Y1(t). So, Z(t)=X1(t)+Y1(t) is a Poisson process. You can of course choose to set p1=p2=1/2.
--------- Original Message ---------
DATE: Sun, 16 Nov 2003 13:09:01
From: Jing Zhang
To: ns-users@ISI.EDU
Cc:
Assume the network traffic arrives at a router during a constant time period D with a average arriving rate \lambda. Its mean is (\lambda)D and its variance is also (\lambda)D.
Can we say this traffic is of Poisson distribution?
I checked the Papoulis book, it says only if certain conditions are met, the moments can determine the corresponding density funciton. These conditions are given in the Moment theorem in Chapter 5.
If the mean and the variance is the same as a Poisson process, can we say this process must be of Poisson distribution?
Thanks in advance.
Jing
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<PRE>M(t10-t0) will be a Poisson variable with rate 10*lambda </PRE><PRE>(similarly M(t20-t0)). However, M(t10-t0)/10 is clearly not a </PRE><PRE>Poisson variable since the ratio may not even be an integer </PRE><PRE>(recall that a Poisson variable must take only integer values).
>From elementary calculations, Var(X1)=lambda/10 and </PRE><PRE>Var(X2)=lambda/20.</PRE><PRE> </PRE>
<BLOCKQUOTE style="PADDING-LEFT: 5px; MARGIN-LEFT: 4px; BORDER-LEFT: #1010ff 3px solid"><FONT face="Times New Roman" size=2>--------- Original Message ---------<BR>
<TABLE cellSpacing=0 cellPadding=0 width=800 border=0>
<TBODY>
<TR>
<TD align=left><B>DATE:</B> Thu, 1 Jan 2004 13:33:22 </TD></TR>
<TR>
<TD align=left><B>From:</B> Jing Zhang <jingz002000@yahoo.com></TD></TR>
<TR>
<TD align=left><B>To:</B> raysaikat@lycos.com</TD></TR>
<TR>
<TD align=left><B>Cc:</B> ns-users@ISI.EDU</TD></TR></TBODY></TABLE></FONT><BR>
<DIV>I have a new case different from what we discussed before. </DIV>
<DIV>Suppose I have a time sequence t0, t1, t2, ..., ti, tj, ...tn, ... and tj - ti = 1 second. The traffic arrival process is a Poisson process along the whole time axis. I define </DIV>
<DIV>X1(t) = M(t10-t0)/10 and X2(t) = M(t20-t0)/20, where M(t10-t0) means the number of arrivals during (t10-t0) period. Then according to my simulation, the variance of X1(t) and X2(t) are different. And both X1(t) and X2(t) are not Poisson processes. Could anybody tell me why X1(t) and X2(t) are not just the same processes as all the individual processes during (tj-ti)? </DIV>
<DIV> </DIV>
<DIV>I appreciate your help.</DIV>
<DIV> </DIV>
<DIV>Jing</DIV>
<DIV> </DIV>
<DIV> </DIV>
<DIV> </DIV>
<DIV><STRONG><EM>Saikat Ray <raysaikat@lycos.com></EM></STRONG> wrote:</DIV>
<BLOCKQUOTE class=replbq style="PADDING-LEFT: 5px; MARGIN-LEFT: 5px; BORDER-LEFT: #1010ff 2px solid"><BR>If you observe a random process for a time period 'T', and the number of arrivals during this period happen to have the same mean and variance, you CANNOT conclude that the process is a Poisson Process. There are many many processes whose mean and variances are independent and they may "happen to be" equal.<BR><BR>The simplest way to ascertain whether a given arrival process is Poisson, is to take a look at its inter-arrival times. If the distribution of the inter-arrival times are exponential (for simulations, you may choose to plot the histogram for a large number of samples, say 10000), as well as they are independent (this part is hard to verify by measurement), then the process is Poisson. But, certainly the first test would be to plot the histogram of the interarrival times and compare with the theoretical pdf.<BR><BR>I do not know whether your current mail is relat!
ed
to your last mail; but I think I should clarify one more thing. As I pointed out last time, if you simply average the cumulative number of arrivals of two independent Poisson processes, the result is NOT a Poisson process. However, if X(t) and Y(t) are two independent Poisson Processes and X1(t) is created from X(t) by sending an arrival at X to X1 with probability 'p1', and similarly for Y1(t); both X1(t) and Y1(t) are Poisson processes on their own right and since X(t) and Y(t) were independent, so are<BR>X1(t) and Y1(t). So, Z(t)=X1(t)+Y1(t) is a Poisson process. You can of course choose to set p1=p2=1/2.<BR><BR><BR><BR>--------- Original Message ---------<BR>DATE: Sun, 16 Nov 2003 13:09:01<BR>From: Jing Zhang <JINGZ002000@YAHOO.COM><BR>To: ns-users@ISI.EDU<BR>Cc: <BR><BR><BR>Assume the network traffic arrives at a router during a constant time period D with a average arriving rate \lambda. Its mean is (\lambda)D and its variance is also (\lambda)D. <BR>Can we say this tr!
aff
ic is of Poisson distribution? <BR><BR>I checked the Papoulis book, it says only if certain conditions are met, the moments can determine the corresponding density funciton. These conditions are given in the Moment theorem in Chapter 5.<BR><BR>If the mean and the variance is the same as a Poisson process, can we say this process must be of Poisson distribution?<BR><BR>Thanks in advance.<BR><BR>Jing<BR><BR><BR>Do you Yahoo!?<BR>Protect your identity with Yahoo! Mail AddressGuard<BR><BR><BR><BR>____________________________________________________________<BR>Enter now for a chance to win a 42" Plasma Television!<BR>http://ad.doubleclick.net/clk;6413623;3807821;f?http://mocda1.com/1/c/563632/113422/313631/313631<BR>AOL users go here: http://ad.doubleclick.net/clk;6413623;3807821;f?http://mocda1.com/1/c/563632/113422/313631/313631<BR>This offer applies to U.S. Residents Only<BR></BLOCKQUOTE>
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