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Stan Pawlukiewicz
Guest
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 Posted:
Thu Dec 15, 2005 5:16 pm Post subject:
Re: Probability |
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abariska@student.ethz.ch wrote:
| Quote: | Jerry Avins write:
...
The probability that a random number lies within some interval is finite
and decreases as the interval gets smaller. In the limit, when the width
of the interval goes to zero -- i.e., becomes a single number -- the
probability [goes to] [becomes] [is] (pick one) zero.
That depends completely on the probability density. If it has a point
mass at that number, then the probability is not equal to zero.
Consider the following probability distribution on the intervall
[0,1], defined by
P[ x \in [0, 1/2[ ] = 1/3
P[ x = 1/2 ] = 1/3
P[ x \in ]1/2, 1] ] = 1/3.
It has positive probability for x being equal to 1/2.
Regards,
Andor
The original post had a plain vanilla iid Gaussian |
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Steve Underwood
Guest
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| Posted:
Thu Dec 15, 2005 5:16 pm Post subject:
Re: Probability |
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Richard Owlett wrote:
| Quote: | Jerry Avins wrote:
Peter K. wrote:
Stan Pawlukiewicz <spam@spam.mitre.org> writes:
All I can say is that it's good that I went to college before I
learned to use C4 in the Army.
You're a scary man, Stan. :-)
Aah, he didn't need C4. I disciplined a teacher in elementary school
with ammonium tri-iodide. (It was fun.)
Jerry
You would be a pair with a kid in my hometown, drove his daschund and
widowed mother nuts with it on the kitchen floor. Local fire department
knew him well -- liked experimenting with hot balloons after summer
drought and acres of long grass downwind.
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Didn't *every* kid wash the blackboard with some ammonium tri-iodide
solution at school? :-\
Steve |
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Guest
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| Posted:
Thu Dec 15, 2005 11:22 pm Post subject:
Re: Probability |
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Hi
I was thinking more on these lines
"If {Xn} is a sequence of i.i.d. random variables with common
characteristic
function F and N is an independent non-negative integer valued random
variable with probability generating function g, then the
characteristic
function of the random sum SN = X1+X2+........+XN is given by g(
F(t))."
and the characteristic funcions determine the distribution. The problem
was to find P(y[n+1] >1) and not P(y[n+1]) =1). Please correct
me if I am thinking wrongly.
I apologize if this seems to be a stupid question
Nithin
Stan Pawlukiewicz wrote:
| Quote: | nithin.pal@gmail.com wrote:
Yup, I agreee....this a tricky problem in Random processes....related
to linear systems, characteristic funtions. This particular operation
would correlate the uncorrelated input signal. So, to find the pdf of
the output is your problem...It has been a long time for me dealing
with this stuff....but I hope my 'clues' are correct
Nithin
P( y(n)=1 ) = 0 , characteristic functions don't change that. Either it
is a typo, or a trick question. |
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Stan Pawlukiewicz
Guest
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| Posted:
Fri Dec 16, 2005 12:02 am Post subject:
Re: Probability |
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within.pal@gmail.com wrote:
| Quote: | Hi
I was thinking more on these lines
"If {Xn} is a sequence of i.i.d. random variables with common
characteristic
function F and N is an independent non-negative integer valued random
variable with probability generating function g, then the
characteristic
function of the random sum SN = X1+X2+........+XN is given by g(
F(t))."
and the characteristic funcions determine the distribution. The problem
was to find P(y[n+1] >1) and not P(y[n+1]) =1). Please correct
me if I am thinking wrongly.
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If you neglect the y(n)=1 part, the easiest solution is given in
Papoulis's book. It's just a integration in x1, x2 of the region x1 + x2
| Quote: | 1, of a 2d uncorrelated Gaussian.
I apologize if this seems to be a stupid question
Nithin
Stan Pawlukiewicz wrote:
nithin.pal@gmail.com wrote:
Yup, I agreee....this a tricky problem in Random processes....related
to linear systems, characteristic funtions. This particular operation
would correlate the uncorrelated input signal. So, to find the pdf of
the output is your problem...It has been a long time for me dealing
with this stuff....but I hope my 'clues' are correct
Nithin
P( y(n)=1 ) = 0 , characteristic functions don't change that. Either it
is a typo, or a trick question.
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Guest
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| Posted:
Fri Dec 16, 2005 8:35 am Post subject:
Re: Probability |
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Stan Pawlukiewicz wrote:
| Quote: | If your in a pinch and not in a mood to argue with your prof, I would
turn in kd_el's answer. It is a cleaver algebraic manipulation. IMHO the
problem is not well posed.
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I don't agree that the problem is not well-posed, and
I don't agree with any of the proposed answers.
We are given a sequence of independent standard
N(0, 1) Gaussian random variables x[n] and a filter
whose output is y[n] = x[n] + x[n-1]. The variance
of y[n] is
var(y[n]) = var(x[n]) + var(x[n-1]) = 2,
and since the x's are gaussian, so is y[n]; that is,
y[n] is a N(0,2) random variable for each choice of
integer n. HOWEVER, y[n] = x[n] + x[n-1] and
y[n+1] = x[n+1] + x[n] are NOT independent
random variables because they both depend
on x[n]. (On the other hand, y[n] and y[n+2] are
indeed independent because they are obtained
from disjoint sets of independent random variables.)
It is easy to show that the covariance of y[n] and
y[n+1] is
cov(y[n], y[n+1]) = cov(x[n] + x[n-1], x[n+1] + x[n])
= cov(x[n], x[n+1]) + cov(x[n], x[n])
+ cov(x[n-1, x[n+1]) + cov(x[n-1, x[n])
= 0 + var(x[n]) + 0 + 0 = var(x[n]) = 1.
Another way of thinking about this is that the
covariance function of the filter output random
process is (1, 2, 1) which is just the autocorrelation
function of the filter impulse response (1, 1).
Thus, y[n] and y[n+1] are N(0,2) random variables
that have the two-dimensional joint Gaussian
distribution with correlation coefficient 1/2. Now,
the question asked of the original poster is
"What is the probability that y[n+1] > 1 given
that y[n] = 1?" This is a standard well-posed
problem in filtering. We observe the filter output
at time n and wish to estimate the filter output
at time n+1. Now, the CONDITIONAL distribution
of y[n+1] given that y[n] was observed to have
value u is itself a Gaussian distribution whose
mean and variance are given by standard formulas
that in this case simplify to u/2 and 3/2 respectively.
Calculation of the conditional probability that
y[n+1] > 1 given that y[n] = 1 is left as an exercise
for the original poster. And no, the answer is
NOT 1/2.
Hope this helps.... |
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Guest
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| Posted:
Fri Dec 16, 2005 9:16 am Post subject:
Re: Probability |
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| Thanks a lot. It sure helps. I am glad to learn something. |
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Stan Pawlukiewicz
Guest
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| Posted:
Fri Dec 16, 2005 5:16 pm Post subject:
Re: Probability |
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dvsarwate@ieee.org wrote:
| Quote: | Stan Pawlukiewicz wrote:
If your in a pinch and not in a mood to argue with your prof, I would
turn in kd_el's answer. It is a cleaver algebraic manipulation. IMHO the
problem is not well posed.
I don't agree that the problem is not well-posed, and
I don't agree with any of the proposed answers.
We are given a sequence of independent standard
N(0, 1) Gaussian random variables x[n] and a filter
whose output is y[n] = x[n] + x[n-1]. The variance
of y[n] is
var(y[n]) = var(x[n]) + var(x[n-1]) = 2,
and since the x's are gaussian, so is y[n]; that is,
y[n] is a N(0,2) random variable for each choice of
integer n. HOWEVER, y[n] = x[n] + x[n-1] and
y[n+1] = x[n+1] + x[n] are NOT independent
random variables because they both depend
on x[n]. (On the other hand, y[n] and y[n+2] are
indeed independent because they are obtained
from disjoint sets of independent random variables.)
It is easy to show that the covariance of y[n] and
y[n+1] is
cov(y[n], y[n+1]) = cov(x[n] + x[n-1], x[n+1] + x[n])
= cov(x[n], x[n+1]) + cov(x[n], x[n])
+ cov(x[n-1, x[n+1]) + cov(x[n-1, x[n])
= 0 + var(x[n]) + 0 + 0 = var(x[n]) = 1.
Another way of thinking about this is that the
covariance function of the filter output random
process is (1, 2, 1) which is just the autocorrelation
function of the filter impulse response (1, 1).
Thus, y[n] and y[n+1] are N(0,2) random variables
that have the two-dimensional joint Gaussian
distribution with correlation coefficient 1/2. Now,
the question asked of the original poster is
"What is the probability that y[n+1] > 1 given
that y[n] = 1?" This is a standard well-posed
problem in filtering.
We observe the filter output
at time n and wish to estimate the filter output
at time n+1. Now, the CONDITIONAL distribution
of y[n+1] given that y[n] was observed to have
value u is itself a Gaussian distribution whose
mean and variance are given by standard formulas
that in this case simplify to u/2 and 3/2 respectively.
Calculation of the conditional probability that
y[n+1] > 1 given that y[n] = 1 is left as an exercise
for the original poster. And no, the answer is
NOT 1/2.
Hope this helps....
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Yeah , your right, I was wrong. I appoligize to the poster and his teacher.
\int_{1}^{\infty} f(x|y=1) dx is the way to go. |
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