Probability Density Function: Aircraft Detection

[jegues]

Homework Statement

See figure attached.

Homework Equations

The Attempt at a Solution

I'm having trouble getting start with this one, but here's what I've got so far.

I assumed R is the signal received by the TDS.

$$P(R=X) = \mu \quad , \quad P(R=N) = 1 - \mu$$

Now in part (a) I believe we are looking for,

$$P( X \geq \gamma A | R = X)$$

The only other thing I can think of applying is Bayes rule,

$$P( X \geq \gamma A | R = X) = \frac{P(X \geq \gamma \cap R = X)}{P(R=X)}$$

Also I think I have to incorporate the probability density function into the problem as well and I haven't done so yet.

Any ideas/suggestions?

Thanks again!

 

[jegues]

I've made a little headway, I will add more work when I get some time to, but here's my idea:

$$R = N + X$$

where,

$$X\in\{ 0 , A \}$$

This simply shifts my Laplace PDF to the right by A units if there is an aircraft, and not otherwise.

So I should be able to describe a conditional PDF based on this.

More work when I return!

 

[haruspex]

In the attachment, X is the received signal. When there is no aircraft, X = N; when there is one, X = A+N. Try to answer (a).

 

[jegues]

EDIT:

$$Y = X + N$$

$$P_{Y|X}(y|X=A) = A + \frac{1}{2}e^{-|y|}$$$$P(Y \geq \gamma A | X =A) = \int_{\gamma A}^{\infty} (A + \frac{1}{2}e^{-|y|})dy$$

Now how do I deal with that integral?

 

[haruspex]

$$Y = X + N$$

That still isn't the assignment of symbols you were given, which is going to be confusing. It is given as:
A = magnitude of the signal, if present
N = noise (random variable)
X = either N or A+N.
If we write S as the r.v. for the signal then S takes the values 0 and A, the latter with probability μ%. X = S+N.

$$P_{Y|X}(y|X=A) = A + \frac{1}{2}e^{-|y|}$$

The LHS expresses a probability, apparently of a specific value of your Y (my X). Since it is a continuous r.v., any specific value has zero probability, but I guess we can think of it as a probability density. The RHS, more seriously, mixes a probability (the second term) with some other type of value (A, the signal strength if present). I guess you meant to write something like P[A+y < Y < A+y+δy|X=A] = δye^-|y|/2. That will change your integral somewhat.

$$P(Y \geq \gamma A | X =A) = \int_{\gamma A}^{\infty} (A + \frac{1}{2}e^{-|y|})dy$$
Now how do I deal with that integral?

Separate the integral into two ranges so that the modulus sign can be evaluated in each.

 

[jegues]

That still isn't the assignment of symbols you were given, which is going to be confusing. It is given as:
A = magnitude of the signal, if present
N = noise (random variable)
X = either N or A+N.
If we write S as the r.v. for the signal then S takes the values 0 and A, the latter with probability μ%. X = S+N.

The LHS expresses a probability, apparently of a specific value of your Y (my X). Since it is a continuous r.v., any specific value has zero probability, but I guess we can think of it as a probability density. The RHS, more seriously, mixes a probability (the second term) with some other type of value (A, the signal strength if present). I guess you meant to write something like P[A+y < Y < A+y+δy|X=A] = δye^-|y|/2. That will change your integral somewhat.

Separate the integral into two ranges so that the modulus sign can be evaluated in each.

I'm still quite confused.

Are you saying my integral is correct or I should be using your expression for the integral?

What I meant by my integral is to show that the Laplace PDF is simply shifted to the right by A when there is an aircraft present, and not shifted at all when there isn't an aircraft present.

What is the δ? There is no symbol given like that in the problem, so I don't know where you got that from.

Can you please clarify?

Thanks again!

 

[haruspex]

Are you saying my integral is correct or I should be using your expression for the integral?

Your integral has to be wrong because the integrand makes no physical sense; it adds a signal level to a probability.
Do you agree that P[A+y < Y < A+y+δy|X=A] = δy e^-|y|/2? So
P[Y < A+y|X=A] = ∫^ye^-|y|/2.dy
P[Y < γA|X=A] = ∫^γA-Ae^-|y|/2.dy
You'll need to consider two cases according to whether γ is more or less than 1.

 

[jegues]

Hello again haruspex,

Your integral has to be wrong because the integrand makes no physical sense; it adds a signal level to a probability.
Do you agree that P[A+y < Y < A+y+δy|X=A] = δy e^-|y|/2? So

Can you explain to me what the δ symbol is as there is no mention of it in the original question. Where is that coming from?

When X = A all that does is shift the pdf of N (given as pN(n) in the question) over to the right by A, correct?

P[Y < A+y|X=A] = ∫^ye^-|y|/2.dy
P[Y < γA|X=A] = ∫^γA-Ae^-|y|/2.dy
You'll need to consider two cases according to whether γ is more or less than 1.

We are told in the question that,

$$0 < \gamma < 1$$

so we only need to consider the one case then, correct?

Also, can you explain the bounds on your integrals? I only see an upper bound and no lower bound.

If I want the probability,

$$P(Y < \gamma A | X = A) = \int _{-\infty}^{\gamma A}\frac{1}{2}e^{-|y-A|}dy$$

I would have written the limits of the integral as such.

Is this incorrect?

Can you explain where you are getting your limits from?

 

[haruspex]

Can you explain to me what the δ symbol is as there is no mention of it in the original question. Where is that coming from?

The δ is implicit in the fact that it is a probability density function. For continuous r.v. X, if you're told the pdf is f(x), that simply means P[x < X < x+δx] → f(x)δx as δx→0.

When X = A all that does is shift the pdf of N (given as pN(n) in the question) over to the right by A, correct?

Yes.

We are told in the question that, $0 < \gamma < 1$ so we only need to consider the one case then, correct?

Yes.

Also, can you explain the bounds on your integrals? I only see an upper bound and no lower bound.

I assumed a lower bound of -∞.

$$P(Y < \gamma A | X = A) = \int _{-\infty}^{\gamma A}\frac{1}{2}e^{-|y-A|}dy$$

That's equivalent to what I wrote.

 

[jegues]

Good then we are on the same page!

So how do I go about simplifying,

$$P(Y < \gamma A | X = A) = \int _{-\infty}^{\gamma A}\frac{1}{2}e^{-|y-A|}dy$$In fact, the probablity I'm looking for in part a) should be this, (or simply 1 - the solution above)

$$P(Y \geq \gamma A | X = A) = \int _{\gamma A}^{\infty}\frac{1}{2}e^{-|y-A|}dy$$

Correct?

But how do we deal with that integral, there is no closed form solution for that, right?

 

[haruspex]

In fact, the probablity I'm looking for in part a) should be this, (or simply 1 - the solution above)

$$P(Y \geq \gamma A | X = A) = \int _{\gamma A}^{\infty}\frac{1}{2}e^{-|y-A|}dy$$

Yes. I switched it to make it conform to the usual cdf form.

But how do we deal with that integral, there is no closed form solution for that, right?

As I said, you have to split the range so that you know how to evaluate |y-A| in each range. When y < A, |y-A| = A-y.