Conditional Probability, Additive Signals

[Anthony45802]

A signal, X, is a random variable with the following density function:
$$f_{X}(x) = \begin{cases} \frac{3}{25}(x-5)^2, & 0 \le x \le 5\\ 0, & otherwise \end{cases}$$

The signal is transmitted through an additive Gaussian noise channel, where the Gaussian noise has a mean of 0 and a variance of 4. The signal and noise are independent.

Find an expression for the conditional density function of the signal, given the observation of the output.

Obviously, this is a homework assignment, so I don't want it done for me; however, I am confused. Perhaps I am just confused by the problem or the wording, but I am totally stuck on what to do.

I believe the output signal should be a convolution where Z = X + Y, and Y is the gaussian(0, 2). After I solve the convolution and receive Z, I don't know what to do.

 

[Stephen Tashi]

My guess is that you need to solve for the conditional cumulative distribution $F(w,y) = P(X + Y \le w | Y = y) = P(X \le w - y | Y = y)$. Then differentiate with respect to w to get the density. (I think you interpreted "gaussian noise channel" correctly in the context of the problem, but I think it has an interpretation as a continuous stochastic process in other contexts.)

 

[Stephen Tashi]

If g(X,Y) is joint density of the convolution X + Y, can you set Y = y and integrate symbolically with respect to X from minus infinity to w-y to obtain $P(X \le w -y| Y = y)$? (I haven't worked the details out, so I don't know.)

 

[ssd]

If output (Z)= noise(Y) + signal (X) then the following is the working principle:
Find the distribution of Z.
Then from the joint distribution Z and X find the distribution X|Z.

 

[blue_raver22]

As a scientist, it is important to understand the concepts and principles behind the problem before attempting to solve it. In this case, the problem involves conditional probability and additive signals, which are fundamental concepts in probability and signal processing, respectively.

Conditional probability refers to the likelihood of an event occurring given that another event has already occurred. In this case, the event of interest is the signal X and the observed output is Z, which is the sum of X and the independent Gaussian noise Y.

To find the conditional density function of X given Z, we can use the Bayes' rule:

f_{X|Z}(x|z) = \frac{f_{Z|X}(z|x)f_X(x)}{f_Z(z)}

Where f_{Z|X}(z|x) is the conditional density function of Z given X, and f_Z(z) is the marginal density function of Z.

Since X and Y are independent, f_{Z|X}(z|x) can be written as the convolution of the individual density functions:

f_{Z|X}(z|x) = f_X(x) * f_Y(z-x)

Substituting this into the Bayes' rule equation, we get:

f_{X|Z}(x|z) = \frac{f_X(x) * f_Y(z-x) * f_X(x)}{f_Z(z)}

Now, we can plug in the given density functions for X and Y:

f_{X|Z}(x|z) = \frac{\frac{3}{25}(x-5)^2 * \frac{1}{\sqrt{8\pi}}e^{-\frac{(z-x)^2}{8}}}{\int_{-\infty}^{\infty}\frac{3}{25}(x-5)^2 * \frac{1}{\sqrt{8\pi}}e^{-\frac{(z-x)^2}{8}} dx}

Simplifying this expression will give us the conditional density function of X given Z.

To solve the convolution and find the marginal density function of Z, you can use the fact that the convolution of two independent Gaussian distributions is also a Gaussian distribution. The mean of the resulting distribution will be the sum of the individual means, and the variance will be the sum of the individual variances.

I hope this helps clarify the problem and guide you in finding the solution. Remember to always understand the concepts and principles behind a problem before attempting