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 October 23rd, 2015, 04:30 AM #1 Newbie   Joined: Jan 2014 Posts: 10 Thanks: 0 Description of distribution w. indicator random variable Dear Forum, I was discussing this very interesting problem with my friend, and I want to hear your opinions on it. Let $\displaystyle X$ be a random variable with unknown distribution, and $\displaystyle c$ be a user-define parameter. Let $\displaystyle A$ be the event that $\displaystyle X = x \leq c$ Define the random variable $\displaystyle Y$, such that $\displaystyle Y = y = g$ if $\displaystyle A$ occurs and $\displaystyle Y = y = 0$ otherwise. How do i give a formal distribution of $\displaystyle Y$, i.e. expected value and variance? If the general case is too difficult, we can assume that $\displaystyle X$ is normally distributed
 October 23rd, 2015, 04:33 PM #2 Global Moderator   Joined: May 2007 Posts: 6,730 Thanks: 689 Since Y has only two possible values (g or 0) it cannot be normally distributed. Since the distribution for X is unknown, you can't say much about the distribution for Y.
October 25th, 2015, 03:57 AM   #3
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Quote:
 Originally Posted by mathman Since Y has only two possible values (g or 0) it cannot be normally distributed. Since the distribution for X is unknown, you can't say much about the distribution for Y.
The OP was saying we can assume X, not Y, is normally distributed... in which case the distribution for X is known.

 October 25th, 2015, 07:12 PM #4 Global Moderator   Joined: May 2007 Posts: 6,730 Thanks: 689 Sorry to have misread the question. The distribution for Y, even if X is normal, will be determined by the mean and variance for X.
 October 26th, 2015, 08:23 AM #5 Senior Member   Joined: Dec 2012 From: Hong Kong Posts: 853 Thanks: 311 Math Focus: Stochastic processes, statistical inference, data mining, computational linguistics If $\displaystyle X \sim N(\mu, \sigma^2)$ and the c-parameter is $\displaystyle \mu$, then $\displaystyle Y \sim Bernoulli(0.5)$ and thus $\displaystyle E(Y) = 0.5$ and $\displaystyle Var(Y) = 0.5(1-0.5) = 0.25$. Not exactly groundbreaking, though :P

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