how to solve the following Probability density function in matlab?
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Hello, i am very new to Matlab and got this question in assignment which i need to submit soon. Any help with coding and a bit explanation would be appreciated.
A probability density function (PDF) is given as follows:
1/4 0<=x<3
f(x) =
5/8-(1/8)x 0<=x<=5
- Plot the PDF in Matlab.
- Calculate and plot the CDF.
- Demonstrate analytically that the mean value associated with the PDF is 49/24.
The answer to the question should be the plots of the PDF and the CDF, as well as the analytical calculation of the mean value.
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Roger Stafford
2014-9-11
There is an obvious misprint in the stated interval. It should be:
1/4 0<=x<3
f(x) =
5/8-(1/8)x 3<=x<=5
As to answering (3.), it all depends on your remembering your calculus enough to be able to integrate a constant times x and a constant times x^2. I think they want this one done by hand. Also you need to know what the definition of "mean value" is.
5 个评论
Roger Stafford
2014-9-17
Your Y is correct up to x = 3. But at that point you have introduced a discontinuity into it. Notice that just before that point your Y was very nearly equal to 3/4, which is correct. Just after it, the value suddenly jumps up to 21/16 (I have corrected your erroneous 1/18*x^2 to 1/16*x^2.) That is because you haven't computed the integration of the PDF properly. First of all, there is a residual 3/4 that is the cumulative probability built up by the prior interval. That can't be suddenly ignored. You need to include it in your Y values. Next, your integration going past there is in error. What you want is the definite integral of 5/8-1/8*t with respect to t from 3 to x, so that is the difference between the two indefinite integrals at x and at 3: (5/8*x-1/16*x^2)-(5/8*3-1/16*3^2). You forgot to include the "-(5/8*3-1/16*3^2)". This is equal to
-21/16+5/8*x-1/16*x^2
This would of course be zero at x = 3, so we have to add in the 3/4 to make it a truly cumulative distribution. The final result would be
-9/16+5/8*x-1/16*x^2
which is what I gave you. Notice that when x = 5, the value of CDF becomes exactly one, which is what it should be.
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Image Analyst
2014-9-11
Try this hint:
x = linspace(0, 5, 800); % Divide section from 0 - 5 up into 800 elements.
fx = 0.25 * ones(1, length(x)); % Initialize fx as all 1/4
% Now need to set the latter part equal to that equation.
latterIndexes = x >= 3;
fx(latterIndexes) = your equation using x(latterIndexes) instead of x.
See how far you get with your homework with this hint.
5 个评论
Image Analyst
2014-9-16
To also plot CDF, use cumsum:
clc; % Clear the command window.
close all; % Close all figures (except those of imtool.)
clear; % Erase all existing variables. Or clearvars if you want.
workspace; % Make sure the workspace panel is showing.
format long g;
format compact;
fontSize = 22;
x = linspace(0, 5, 800); % Divide section from 0 - 5 up into 800 elements.
fx = 0.25 * ones(1, length(x)); % Initialize fx as all 1/4
% Now need to set the latter part equal to that equation.
latterIndexes = x >= 3;
fx(latterIndexes) = 5/8 - (1/8)*x(latterIndexes);
% Now we're done creating the function.
% All that's left to do is plot it.
subplot(1,2,1);
plot(x, fx, 'b-', 'LineWidth', 3);
ylim([0,0.3]);
grid on;
xlabel('x', 'FontSize', 25);
ylabel('fx', 'FontSize', 25);
title('PDF', 'FontSize', fontSize);
% Compute CDF from the PDF, fx.
cdf = cumsum(fx);
% Convert to percentage
cdf = 100 * cdf / cdf(end);
subplot(1,2,2);
plot(x, cdf, 'b-', 'LineWidth', 3);
ylim([0, 100]);
grid on;
xlabel('x', 'FontSize', 25);
ylabel('Percentage', 'FontSize', 25);
title('CDF', 'FontSize', fontSize);
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/181158/image.jpeg)
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