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John Agkali
0

Continuous laplacian random variable

Asked by John Agkali
on 29 May 2013

I want to plot the outcomes of a Laplacian random Variable when i have the type : A>=sqrt(sigma^2/2)*ln(1/0.01) with sigma=1.I understand what i need to plot but i can not understand that how can I present discrete values for 50 trials ..?Could anyone suggest me how can I do it ?

Thanks!

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1 Answer

Answer by Tom Lane
on 30 May 2013
 Accepted answer

Take a look at this:

sigma = 1;
p = rand(10000,1);
A=sqrt(sigma^2/2)*log(1./p);
hist(A,100)

I'm not sure if it's what you want. It uses random P in place of the 0.1 value from your expression. Otherwise it uses that expression to generate random numbers, and makes a histogram of them. Notice that I use "./" to perform division on every element of P.

Perhaps you can figure out how to present the result of 50 trials by looking at this.

  8 Comments

Tom Lane
on 5 Jun 2013

You need to find some way to get either a positive or a negative sign. This is one way. It produces the result +1 when its input is positive, and -1 when its input is negative.

The first code segment that you gave me, the one with function hist in it, could you briefly tell me what a histogramm represents ? How can i "decode" its meaning ? And for what purposes is used ?

For example,if I hist the function in the second code segment that you sent me, hist(A), what can I describe about that ? Is there something very important about histogramm I should know ?

Thank you!!

Tom Lane
on 9 Jun 2013

The image you posted showed each trial as a dot, plotted with a line connecting it to zero. That works fine with the 50 or so trials you have.

I had lots of trials so I made a histogram. It counts the number of trials in each of several bins, and draws a bar of that height. So the heights of the bars give a picture of the distribution of points across bins. It may be this is not important for what you are trying to do.


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