Poisson random process problem

[ashah99]

Homework Statement

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Relevant Equations

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Hello all, sorry for the large wall of text but I'm really trying to understanding a problem from a study guide. I am quite unsure on how to approach the following multi-part problem. Any help would be appreciated.

Problem:

Useful references I'm using to attempt the problem

My attempt:
For part (a), my reasoning is the following, and if anyone can kindly check on these answers, that would be great.

(i) Poisson(5), since this describes the number of arrivals (hits) in a window, so the RV should be Poisson(lambda*T), where λ = 1 photon/us and T is 5 us duration, thus Poisson(5)
(ii) Poisson(2), Z is the number of events in [5,7], so this is Poisson(7-5) = Poisson(2)
(III) Exp(1), since waiting times of a Poisson process are i.i.d. Exponential with parameter λ = 1
(iv) Erlang(1,2) since this is the waiting time for the kth arrival

Based off the info above, then for parts (b) and (c) I get the following:

 

[Office_Shredder]

(i) is just a distribution that says if X is nonzero, it's not asking for the number of hits.

For (ii), what does that parameter (2) mean?
You should think about this more carefully.

I didn't check the rest.

 

[ashah99]

(i) is just a distribution that says if X is nonzero, it's not asking for the number of hits.

For (ii), what does that parameter (2) mean?
You should think about this more carefully.

I didn't check the rest.

My mistake on (i)
X is either non-zero or it isn't. It's non-zero with probability p=1-e^(-5)
Thus it's Bernoulli with parameter p -> X is Bern( 1-e^(-5) ) should be the correct answer I hope

(ii) Z is the number of events in [5,7], since X and Y overlap by 5 us, so Z is Poisson(2)

 

[Office_Shredder]

Sorry I misread the problem, your answer for (ii) is fine

(iii) is problematic, though I suspect your answer is correct. The expected value is 1 microsecond, and it never says as far as I can tell what unit of time the measurement is in (if it' in seconds, the answer is exp(100000))

(iv) ooks fine as well though Wikipedia at least things that k comes first in the two parameters.

For (b) I agree with your answer, but I think you should meditate on it a bit and understand how you can construct it without computing any square roots.

(C) looks wrong to me. First, there should be a Y in your formula, not a $x$. More importantly though if $Y=7$ then your formula says the expected value for $X$ is 7 also, which seems impossible :)

 

[ashah99]

Sorry I misread the problem, your answer for (ii) is fine

(iii) is problematic, though I suspect your answer is correct. The expected value is 1 microsecond, and it never says as far as I can tell what unit of time the measurement is in (if it' in seconds, the answer is exp(100000))

(iv) ooks fine as well though Wikipedia at least things that k comes first in the two parameters.

For (b) I agree with your answer, but I think you should meditate on it a bit and understand how you can construct it without computing any square roots.

(C) looks wrong to me. First, there should be a Y in your formula, not a $x$. More importantly though if $Y=7$ then your formula says the expected value for $X$ is 7 also, which seems impossible :)

Good point on part (iii) with the units. Hmm I would note down both for my studies
Part (c) is an error on my part due to copy/pasting. You are correct in that formula could contain a $y$
I believe the formula should also say $u_x$ so if I make those corrections, I get a final answer of $E(X |Y=y) = y - 2$ I believe.

Would you agree here?

 

[Office_Shredder]

I don't like that answer either. What if $y=1$?

 

[ashah99]

I don't like that answer either. What if $y=1$?

Hmm that would be problematic resulting in the expectation being negative. Do you a better approach?

 

[Office_Shredder]

Can you try one more time to Carefully write down all the parameters you needed to compute your mmse?

If you want the clever way to do this, that really doesn't scale at all to any other example of this kind of problem (except conveniently gaussians because the sum of gaussian is also a gaussian)

Let $X_1,X_2,...,X_7$ be independent poisson(1) random variables. Then we can pretend $X$ is the sum of the first 5, and $Y$ is the sum of the first seven. Given that all seven of these identical distributions sum to $Y$, what's your best guess for the sum of the first five?

 

[ashah99]

Can you try one more time to Carefully write down all the parameters you needed to compute your mmse?

If you want the clever way to do this, that really doesn't scale at all to any other example of this kind of problem (except conveniently gaussians because the sum of gaussian is also a gaussian)

Let $X_1,X_2,...,X_7$ be independent poisson(1) random variables. Then we can pretend $X$ is the sum of the first 5, and $Y$ is the sum of the first seven. Given that all seven of these identical distributions sum to $Y$, what's your best guess for the sum of the first five?

The parameters I used were $µ_X = 5, µ_Y = 7, σ_X = √5, σ_Y = √7, and ρ = √(5/7)$
Using the formula above, I get
$E(X | Y = y) =µ_X +ρσ_X( (Y-µ_Y ) / √7) = 5 + √(5/7)√5 ( (y-7 ) /√7)$
and then $E(X | Y = y) = y - 2$ but that turns out to be incorrect.

 

[Office_Shredder]

You need to do your algebra more carefully.

$x=5+\frac{\sqrt{5}\sqrt{5}}{\sqrt{7}\sqrt{7}}(y-7)$
$x=5+\frac{5}{7}(y-7)$
$x=5+\frac{5}{7}y-5$
$x=\frac{5}{7}y$

 

[ashah99]

You need to do your algebra more carefully.

$x=5+\frac{\sqrt{5}\sqrt{5}}{\sqrt{7}\sqrt{7}}(y-7)$
$x=5+\frac{5}{7}(y-7)$
$x=5+\frac{5}{7}y-5$
$x=\frac{5}{7}y$

Thank you for your help. I realize the mistakes now. I had the formula wrong originally by using the mean of y in stead of mean of x hence had the extra factor of +2. So yes the answer now makes sense: when y=7, then x=5 which is expected.