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dbell
Guest
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 Posted:
Thu Dec 15, 2005 1:17 am Post subject:
Re: Probability |
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And the practical application (besides getting credit) is?
Dirk
ngeva0 wrote:
| Quote: | x[n] is a zero-mean white Gaussian random process with variance of 1 and
y[n] is the output when x[n] is filtered using a two-tap FIR filter with
coefficients of [1 1]. What's the probability of the event {y[n+1]>1}
given that y[n]=1? |
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Jerry Avins
Guest
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| Posted:
Thu Dec 15, 2005 1:17 am Post subject:
Re: Probability |
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ngeva0 wrote:
| Quote: | y[n] = 1
=> x[n-1] + x[n] = 1
=> x[n] = 1 - x[n-1]
P{ y[n+1]>1 }
= P{ (x[n+1]+x[n]) > 1 }
= P{ (x[n+1]+1-x[n-1]) > 1 }
= P{ (x[n+1]-x[n-1]) > 0 }
= P{ x[n+1]>x[n-1] }
Since x[n] has a symmetrical distribution
P{ x[n+1]>x[n-1] } = P{ x[n+1]<x[n-1] } = 1/2
ngeva0 wrote:
x[n] is a zero-mean white Gaussian random process with variance of 1
and
y[n] is the output when x[n] is filtered using a two-tap FIR filter
with
coefficients of [1 1]. What's the probability of the event {y[n+1]>1}
given that y[n]=1?
Since P( y(n) = 1 ) = 0, I think P( y(n+1) > 1 ) should be a very small
number?
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{p(y[n]) = x} = 0 for all x. It seems paradoxical at first, but it's
true. Since it must equal something, that just shows that the sum of an
infinite number of infinitesimals can be non zero.
Jerry
--
Engineering is the art of making what you want from things you can get.
ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ |
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Stan Pawlukiewicz
Guest
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| Posted:
Thu Dec 15, 2005 1:17 am Post subject:
Re: Probability |
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ngeva0 wrote:
| Quote: | y[n] = 1
=> x[n-1] + x[n] = 1
=> x[n] = 1 - x[n-1]
P{ y[n+1]>1 }
= P{ (x[n+1]+x[n]) > 1 }
= P{ (x[n+1]+1-x[n-1]) > 1 }
= P{ (x[n+1]-x[n-1]) > 0 }
= P{ x[n+1]>x[n-1] }
Since x[n] has a symmetrical distribution
P{ x[n+1]>x[n-1] } = P{ x[n+1]<x[n-1] } = 1/2
ngeva0 wrote:
x[n] is a zero-mean white Gaussian random process with variance of 1
and
y[n] is the output when x[n] is filtered using a two-tap FIR filter
with
coefficients of [1 1]. What's the probability of the event {y[n+1]>1}
given that y[n]=1?
Since P( y(n) = 1 ) = 0, I think P( y(n+1) > 1 ) should be a very small
number?
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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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Tim Wescott
Guest
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| Posted:
Thu Dec 15, 2005 1:17 am Post subject:
Re: Probability homework? |
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Jerry Avins wrote:
| Quote: | ngeva0 wrote:
ngeva0 wrote:
x[n] is a zero-mean white Gaussian random process with variance of 1
and
y[n] is the output when x[n] is filtered using a two-tap FIR filter
with
coefficients of [1 1]. What's the probability of the event {y[n+1]>1}
given that y[n]=1?
Is the noise bandlimited before filtering? What is the effect of your
filter on any signal? (Given y[n] = a and y[n+1] = b, what will the
filter output be?)
Jerry
--
Engineering is the art of making what you want from things you can get.
ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ
the output y[n]=x[n]+x[n-1] after filtering. therefore,
y[n+1]=x[n+1]+x[n]
Since the input to the filter isn't bandlimited, there will be aliasing
at any sample rate, so beware of intuitive results. Nevertheless, if
y[n] = 1, then x[n] + x[n-1] = n. You need to calculate the probability
that x[n+1] will exceed x[n-1]. That's not trivial, but at least it's
not a conditional probability. Pick a (Gaussian) number. Pick another.
What is the probability that the second is more positive than the first?
Jerry
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When talking of sampled data "white" means that each sample is
independent of the next -- it does _not_ mean that you've sampled white
noise. This is good, because truly white continuous-time noise would
have infinite power, and that's impossible.
--
Tim Wescott
Wescott Design Services
http://www.wescottdesign.com |
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Carlos Moreno
Guest
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| Posted:
Thu Dec 15, 2005 1:17 am Post subject:
Re: Probability |
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Jerry Avins wrote:
| Quote: | {p(y[n]) = x} = 0 for all x. It seems paradoxical at first, but it's
true. Since it must equal something, that just shows that the sum of an
infinite number of infinitesimals can be non zero.
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<Nitpick pedantic_level="50%">
No, not really. It just shows that "infinite/infinity" is a
mathematical abstraction quite different than the mathematical
abstraction that we call numbers (at least real numbers) --
arithmetic rules for numbers *do not* apply to infinity (no
matter how much we try and how much we try to convince
ourselves that it does)
P(y[n] = X) is 0 because the probability is defined as a
*limit*; it's not an infinitesimal number: it is *zero*;
*exactly zero*, not "similar to zero", not "like zero",
no: *really and exactly zero*. You add as many of those
as you want, and you still get zero. You take a interval
of non-finite size, and then you get something different
than zero. Not because you added an infinite amount of
zeros -- there is no such thing as that.
The thing is, P(X < y[n] < X+epsilon) is equal to the
fraction epsilon divided by the range of possible values,
if we're talking about a uniform distribution. P(y[n] = X)
is nothing else than the limit as epsilon approaches 0
of the above -- that limit is zero. (the reasoning has
to be extended to the general case of an arbitrary
probability function, but the argument goes more or less
the same way)
</Nitpick>
Yes, I know this may actually be more confusing for the
OP... (sorry! :-))
Cheers,
Carlos
-- |
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Jerry Avins
Guest
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| Posted:
Thu Dec 15, 2005 1:17 am Post subject:
Re: Probability |
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Carlos Moreno wrote:
| Quote: | Jerry Avins wrote:
{p(y[n]) = x} = 0 for all x. It seems paradoxical at first, but it's
true. Since it must equal something, that just shows that the sum of
an infinite number of infinitesimals can be non zero.
Nitpick pedantic_level="50%"
No, not really. It just shows that "infinite/infinity" is a
mathematical abstraction quite different than the mathematical
abstraction that we call numbers (at least real numbers) --
arithmetic rules for numbers *do not* apply to infinity (no
matter how much we try and how much we try to convince
ourselves that it does)
P(y[n] = X) is 0 because the probability is defined as a
*limit*; it's not an infinitesimal number: it is *zero*;
*exactly zero*, not "similar to zero", not "like zero",
no: *really and exactly zero*. You add as many of those
as you want, and you still get zero. You take a interval
of non-finite size, and then you get something different
than zero. Not because you added an infinite amount of
zeros -- there is no such thing as that.
The thing is, P(X < y[n] < X+epsilon) is equal to the
fraction epsilon divided by the range of possible values,
if we're talking about a uniform distribution. P(y[n] = X)
is nothing else than the limit as epsilon approaches 0
of the above -- that limit is zero. (the reasoning has
to be extended to the general case of an arbitrary
probability function, but the argument goes more or less
the same way)
/Nitpick
Yes, I know this may actually be more confusing for the
OP... (sorry! :-))
|
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.
Jerry
--
Engineering is the art of making what you want from things you can get.
ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ |
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Guest
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| Posted:
Thu Dec 15, 2005 1:17 am Post subject:
Re: Probability |
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Maybe the better way to say this is:
Since x[n+1] and x[n-1] are i.i.d. random variables with Gaussian
distribution
P{ x[n+1]>x[n-1] } = P{ x[n+1]<x[n-1] } = 1/2
Symmetry of the distribution is not necessary. I added the word
"Gaussian" to avoid the case mentioned by Andor.
| Quote: | Since x[n] has a symmetrical distribution
P{ x[n+1]>x[n-1] } = P{ x[n+1]<x[n-1] } = 1/2 |
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Guest
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| Posted:
Thu Dec 15, 2005 1:17 am Post subject:
Re: Probability |
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Maybe the better way to say this is:
Since x[n+1] and x[n-1] are both i.i.d. random variables with Gaussian
distribution
P{ x[n+1]>x[n-1] } = P{ x[n+1]<x[n-1] } = 1/2
(Symmetry of the distribution is not necessary. I added the word
"Gaussian distribution" to avoid the case mentioned by Andor.)
| Quote: | Since x[n] has a symmetrical distribution
P{ x[n+1]>x[n-1] } = P{ x[n+1]<x[n-1] } = 1/2 |
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ngeva0
Guest
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| Posted:
Thu Dec 15, 2005 1:17 am Post subject:
Re: Probability |
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| Quote: | y[n] = 1
=> x[n-1] + x[n] = 1
=> x[n] = 1 - x[n-1]
P{ y[n+1]>1 }
= P{ (x[n+1]+x[n]) > 1 }
= P{ (x[n+1]+1-x[n-1]) > 1 }
= P{ (x[n+1]-x[n-1]) > 0 }
= P{ x[n+1]>x[n-1] }
Since x[n] has a symmetrical distribution
P{ x[n+1]>x[n-1] } = P{ x[n+1]<x[n-1] } = 1/2
ngeva0 wrote:
x[n] is a zero-mean white Gaussian random process with variance of
and
y[n] is the output when x[n] is filtered using a two-tap FIR filte
with
coefficients of [1 1]. What's the probability of the event {y[n+1]>1}
given that y[n]=1?
|
Since P( y(n) = 1 ) = 0, I think P( y(n+1) > 1 ) should be a very smal
number? |
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Peter K.
Guest
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| Posted:
Thu Dec 15, 2005 8:08 am Post subject:
Re: Probability |
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Stan Pawlukiewicz <spam@spam.mitre.org> writes:
| Quote: | All I can say is that it's good that I went to college before I
learned to use C4 in the Army.
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You're a scary man, Stan. :-)
Ciao,
Peter K. |
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Jerry Avins
Guest
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| Posted:
Thu Dec 15, 2005 9:12 am Post subject:
Re: Probability |
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Peter K. wrote:
| Quote: | 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. :-)
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Aah, he didn't need C4. I disciplined a teacher in elementary school
with ammonium tri-iodide. (It was fun.)
Jerry
--
Engineering is the art of making what you want from things you can get.
ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ |
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Jerry Avins
Guest
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| Posted:
Thu Dec 15, 2005 9:15 am 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.
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AWGN: continuous, with continuous derivitives of all orders.
Your exception noted. Thanks.
Jerry
--
Engineering is the art of making what you want from things you can get.
ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ |
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Jerry Avins
Guest
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| Posted:
Thu Dec 15, 2005 9:15 am Post subject:
Re: Probability |
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kd_ei@yahoo.com wrote:
| Quote: | Maybe the better way to say this is:
Since x[n+1] and x[n-1] are both i.i.d. random variables with Gaussian
distribution
P{ x[n+1]>x[n-1] } = P{ x[n+1]<x[n-1] } = 1/2
(Symmetry of the distribution is not necessary. I added the word
"Gaussian distribution" to avoid the case mentioned by Andor.)
|
I don't see that. Zero mean. If x[n-1] is 5 sigmas positive, it's
certainly not true that x[n+1] is as likely to be larger as smaller. Now
it's perfectly possible that x[n-1] is a large negative number offset by
x[n]. Bit the sum being 1 introduces a bias that needs to be accounted
for or shown to be irrelevant.
Jerry
--
Engineering is the art of making what you want from things you can get.
ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ |
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Jerry Avins
Guest
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| Posted:
Thu Dec 15, 2005 5:16 pm Post subject:
How to cure (at least tame) a sick teacher. (was Probability |
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Steve Underwood wrote:
| Quote: | Richard Owlett wrote:
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.
Didn't *every* kid wash the blackboard with some ammonium tri-iodide
solution at school? :-\
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This wasn't the blackboard. The teacher would slam the door whenever she
got frustrated. She seemed to be telling us "Nyah nyah! I can get away
with it but you can't." I came in one Monday morning with an eyedropper
of water and iodide crystals and squirted it into the lock. I went home
for lunch, then did the same. Just before leaving for the day, another
dropper of plain water. Same routine Tue - Fri, leaving the weekend for
it to dry out. (Before leaving Friday, I loosened the setscrews in the
knobs instead of adding water.) Next Monday, when she slammed the door,
it was louder than usual and the knob blew off. She panicked when she
couldn't get the door open; after a time someone opened it with her
letter opener.
Jerry
--
Engineering is the art of making what you want from things you can get.
ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ |
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Richard Owlett
Guest
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| Posted:
Thu Dec 15, 2005 5:16 pm Post subject:
Re: Probability |
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Jerry Avins wrote:
| Quote: | 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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