To compute probabilities for an exponential, Rayleigh, and Weibull random variable using MATLAB, we’ll utilize both the PDF<\/strong> (Probability Density Function) and CDF<\/strong> (Cumulative Distribution Function) functions for each distribution. I’ll guide you on how to do this step-by-step and explain the expected results.<\/p>\n\n\n\n The Exponential Distribution<\/strong> is used to model the time between events in a Poisson process. The PDF for an exponential distribution with rate parameter \u03bb\\lambda\u03bb is:fX(x)=\u03bbe\u2212\u03bbxfor x\u22650f_X(x) = \\lambda e^{-\\lambda x} \\quad \\text{for } x \\geq 0fX\u200b(x)=\u03bbe\u2212\u03bbxfor x\u22650<\/p>\n\n\n\n The CDF is:FX(x)=1\u2212e\u2212\u03bbxF_X(x) = 1 – e^{-\\lambda x}FX\u200b(x)=1\u2212e\u2212\u03bbx<\/p>\n\n\n\n MATLAB Code for PDF and CDF (Exponential Distribution)<\/strong>:<\/p>\n\n\n\n Expected Probabilities for Exponential Distribution<\/strong>:<\/p>\n\n\n\n The theoretical probabilities can also be calculated directly from the CDF:<\/p>\n\n\n\n The Rayleigh Distribution<\/strong> is a special case of the Weibull distribution. Its PDF is:fX(x)=x\u03c32e\u2212x22\u03c32for x\u22650f_X(x) = \\frac{x}{\\sigma^2} e^{-\\frac{x^2}{2\\sigma^2}} \\quad \\text{for } x \\geq 0fX\u200b(x)=\u03c32x\u200be\u22122\u03c32×2\u200bfor x\u22650<\/p>\n\n\n\n The CDF is:FX(x)=1\u2212e\u2212x22\u03c32F_X(x) = 1 – e^{-\\frac{x^2}{2\\sigma^2}}FX\u200b(x)=1\u2212e\u22122\u03c32×2\u200b<\/p>\n\n\n\n MATLAB Code for PDF and CDF (Rayleigh Distribution)<\/strong>:<\/p>\n\n\n\n You can compute the probabilities using a similar loop and CDF method.<\/p>\n\n\n\n The Weibull Distribution<\/strong> has the following PDF:fX(x)=k\u03bb(x\u03bb)k\u22121e\u2212(x\/\u03bb)kf_X(x) = \\frac{k}{\\lambda} \\left( \\frac{x}{\\lambda} \\right)^{k-1} e^{-(x\/\\lambda)^k}fX\u200b(x)=\u03bbk\u200b(\u03bbx\u200b)k\u22121e\u2212(x\/\u03bb)k<\/p>\n\n\n\n The CDF is:FX(x)=1\u2212e\u2212(x\/\u03bb)kF_X(x) = 1 – e^{-(x\/\\lambda)^k}FX\u200b(x)=1\u2212e\u2212(x\/\u03bb)k<\/p>\n\n\n\n MATLAB Code for PDF and CDF (Weibull Distribution)<\/strong>:<\/p>\n\n\n\n Theoretically, the probabilities computed from both the PDF and CDF should match because the CDF represents the cumulative sum of the probability density over the interval, and the PDF represents the continuous probability at any point.<\/p>\n\n\n\n However, in practice, discretization errors<\/strong> can occur when using numerical methods (like summing the PDF in a loop). The integration of the PDF to obtain the CDF may be approximated differently depending on the step size (the resolution of the X values). The choice of interval step size ( This approach will allow you to examine the behavior of the exponential, Rayleigh, and Weibull distributions, and understand the practical limitations of numerical integration.<\/p>\n\n\n\n Use Matlab’s “pdf’ function to create the pdfs for an exponential an exponential random variable X_ Using single for-loop , compute the following probabilities: P[O < X < 1] P[O < X < 5] PIX > 6] Do the same for when X is a Rayleigh and Weibull random variable_ Compute the same probabilities using […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-239238","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/239238","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=239238"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/239238\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=239238"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=239238"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=239238"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}1. Exponential Distribution<\/strong><\/h3>\n\n\n\n
matlabCopyEdit
lambda = 1; % Rate parameter\nX = 0:0.01:10; % Range of X\n\n% PDF using the pdf function\npdf_exp = lambda * exp(-lambda * X);\n\n% CDF using the cdf function\ncdf_exp = 1 - exp(-lambda * X);\n\n% Using a loop to compute probabilities\nP_0_1 = sum(pdf_exp(X > 0 & X < 1)) * 0.01;\nP_0_5 = sum(pdf_exp(X > 0 & X < 5)) * 0.01;\nP_6_inf = sum(pdf_exp(X > 6)) * 0.01;\n<\/code><\/pre>\n\n\n\n\n
matlabCopyEdit
P_0_1_theory = cdf_exp(find(X == 1)) - cdf_exp(find(X == 0));\nP_0_5_theory = cdf_exp(find(X == 5)) - cdf_exp(find(X == 0));\nP_6_inf_theory = 1 - cdf_exp(find(X == 6));\n<\/code><\/pre>\n\n\n\n2. Rayleigh Distribution<\/strong><\/h3>\n\n\n\n
matlabCopyEdit
sigma = 1; % Parameter for Rayleigh distribution\npdf_rayleigh = (X \/ sigma^2) .* exp(-X.^2 \/ (2 * sigma^2));\ncdf_rayleigh = 1 - exp(-X.^2 \/ (2 * sigma^2));\n<\/code><\/pre>\n\n\n\n3. Weibull Distribution<\/strong><\/h3>\n\n\n\n
matlabCopyEdit
k = 1.5; % Shape parameter\nlambda = 1; % Scale parameter\npdf_weibull = (k \/ lambda) * (X \/ lambda).^(k - 1) .* exp(-(X \/ lambda).^k);\ncdf_weibull = 1 - exp(-(X \/ lambda).^k);\n<\/code><\/pre>\n\n\n\n4. Why PDF and CDF May Not Be Exactly the Same<\/strong><\/h3>\n\n\n\n
0.01<\/code> in this case) plays a crucial role in the accuracy of the results. If the step size is too large, you may miss important changes in the probability, which can lead to slight discrepancies between the PDF-based and CDF-based calculations.<\/p>\n\n\n\n
\n\n\n\n5. Summary of Steps<\/strong><\/h3>\n\n\n\n
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pdf<\/code> and cdf<\/code>).<\/li>\n\n\n\n![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-12.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"