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 February 2nd, 2016, 12:45 PM #1 Newbie   Joined: Feb 2016 From: Michigan Posts: 2 Thanks: 0 Random variable with continuous distribution Hello, I'm stuck with following problem: Random variable X with the continuous distribution for some arbitrary constant a > 0: $\displaystyle p(x) = 3/4a³(a² - x²)$ if |x| < a otherwise 0 i. Draw the distribution ii. Show that p(x) is a valid distribution Some hints how to get started with the problem would be appreciated! February 2nd, 2016, 02:49 PM #2 Math Team   Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244 Can you clarify $p(x)$? Is it $\dfrac{3}{4a^2(a^2 - x^2)}$, $\dfrac{3}{4a^2}(a^2 - x^2)$, $\dfrac{3}{4}a^2(a^2 - x^2)$? A small latex tip; If you want to make your exponents more visible, use ^n. For example, the code a^n will give $a^n$. Thanks from Prob February 2nd, 2016, 04:41 PM   #3
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 Originally Posted by Prob Hello, I'm stuck with following problem: Random variable X with the continuous distribution for some arbitrary constant a > 0: $\displaystyle p(x) = 3/4a³(a² - x²)$ if |x| < a otherwise 0 i. Draw the distribution ii. Show that p(x) is a valid distribution Some hints how to get started with the problem would be appreciated!
For i), just substitute, say, a = 1 and plot some points. (Or use use the first and second derivative tests - your choice lol.)

For ii), you have to show that the area under the curve = 1, i.e. integrate p(x) from -a to a. February 2nd, 2016, 09:25 PM #4 Newbie   Joined: Feb 2016 From: Michigan Posts: 2 Thanks: 0 p(x) is: $\displaystyle p(x) = \frac{3}{4a^{3}}(a^2-x^2)$ Tags continuous, distribution, random, variable Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post lstevens Probability and Statistics 1 October 19th, 2015 01:39 PM tayyeb Probability and Statistics 6 March 19th, 2015 11:50 AM rsashwinkumar Advanced Statistics 3 January 7th, 2013 03:25 PM fin0c Advanced Statistics 1 January 23rd, 2011 10:43 PM alfangio Advanced Statistics 1 March 17th, 2009 01:46 PM

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