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	<title>Comments on: Solving exponential distribution probability using Calculus</title>
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	<description>Ramblings of a Geek</description>
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		<title>By: nikola</title>
		<link>http://www.jeremyjohnstone.com/blog/2010-02-01-solving-exponential-distribution-probability-using-calculus.html/comment-page-1#comment-1027</link>
		<dc:creator>nikola</dc:creator>
		<pubDate>Thu, 11 Nov 2010 22:59:16 +0000</pubDate>
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		<description>By the way - here&#039;s a link to some statistical simulation code ( written in R ) dealing with the problem for survival and failure and estimation of Distribution parameters from such data. https://github.com/chochkov/mle_estimation. There&#039;s also Latex and Pdf files in case anyone would be interested..</description>
		<content:encoded><![CDATA[<p>By the way &#8211; here&#8217;s a link to some statistical simulation code ( written in R ) dealing with the problem for survival and failure and estimation of Distribution parameters from such data. <a href="https://github.com/chochkov/mle_estimation" rel="nofollow">https://github.com/chochkov/mle_estimation</a>. There&#8217;s also Latex and Pdf files in case anyone would be interested..</p>
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		<title>By: nikola</title>
		<link>http://www.jeremyjohnstone.com/blog/2010-02-01-solving-exponential-distribution-probability-using-calculus.html/comment-page-1#comment-1008</link>
		<dc:creator>nikola</dc:creator>
		<pubDate>Tue, 09 Nov 2010 16:16:43 +0000</pubDate>
		<guid isPermaLink="false">http://www.jeremyjohnstone.com/?p=1542#comment-1008</guid>
		<description>nice exposition :) you should maybe mention that the two variables are independent though to avoid confusion and for P(AB) = P(A).P(B) to be applicable..

Also, another interesting question is - what&#039;s the probability that variable X2 &#039;survives&#039; certain x while X1 fails. In this case you will have p = pdfX1 * (1 - CDFX2)</description>
		<content:encoded><![CDATA[<p>nice exposition <img src='http://www.jeremyjohnstone.com/wordpress/wp-includes/images/smilies/icon_smile.gif' alt=':)' class='wp-smiley' />  you should maybe mention that the two variables are independent though to avoid confusion and for P(AB) = P(A).P(B) to be applicable..</p>
<p>Also, another interesting question is &#8211; what&#8217;s the probability that variable X2 &#8217;survives&#8217; certain x while X1 fails. In this case you will have p = pdfX1 * (1 &#8211; CDFX2)</p>
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