Probablility density function (randon) variable (Help!!!!)

I have been trying to solve this question for the past 2 day without no success.. I wonder if you guys can help me out...

question) Let x be the transmitted signal and Y the received signal the condition density of Y given X is as follows:

f (y/x) = k exp(- |y-x| /2)
where, k is constant, -inf < y < +inf , and X is uniformily distribuited on [0, 1].

1) Find k ?
2) Find f?

I find k to be 1/4 but I am no longer sure of the value, fffffffffffffffff

Re: Probablility density function (randon) variable (Help!!!

Your problem comes from a simple model y=x+z, where z has a probability density function with the form

f(z)=kexp(-|z|/2)

and the underlying assumption is that x and z are two independent random variables, you will get the resulting probability density function for y as you stated.

With above discussion, it is straight forward to show

(1) k=1/2, based on the fact that integral of f(z) has to be unity in order for it to be a valid probability density function. Expression for f(z) follows immediately.

(2) since y=x+z, and x & z are independent, you can use the well-known result f=f(x) * f(z), where * denotes the convolution operator, to find the expression for f.

Extra: (a) the term z is often called Laplacian noise. It is a noise model often used to model the interference term in UWB system.

Hope that helps!
PPF

angolaX said:

I have been trying to solve this question for the past 2 day without no success.. I wonder if you guys can help me out...

question) Let x be the transmitted signal and Y the received signal the condition density of Y given X is as follows:

f (y/x) = k exp(- |y-x| /2)
where, k is constant, -inf < y < +inf , and X is uniformily distribuited on [0, 1].

1) Find k ?
2) Find f?

I find k to be 1/4 but I am no longer sure of the value, fffffffffffffffff

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