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Continuous laplacian random variable

Asked by John Agkali

John Agkali (view profile)

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 ?



John Agkali

John Agkali (view profile)


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1 Answer

Answer by Tom Lane

Tom Lane (view profile)

on 30 May 2013
Accepted answer

Take a look at this:

sigma = 1;
p = rand(10000,1);

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.


Tom Lane

Tom Lane (view profile)

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.

John Agkali

John Agkali (view profile)

on 8 Jun 2013

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

Tom Lane (view profile)

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.

Tom Lane

Tom Lane (view profile)

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