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I am looking for the mean and variance of a bernoulli distribution of two normally distributed random variables.

An example would be as follows: A coin flip

  • heads: draw from a $N(\mu1,\sigma1)$

  • tails: draw from a $N(\mu2,\sigma2)$

For the expected value of the coinflip I would think it equals $p*\mu1+(1-p)*\mu2$. Where $p$ is 0.5 in this case.

But I have no idea how to compute the variance of this bernoulli distribution.

Could somebody help me with this?

Piet Voerman
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  • There is no such a thing as "Bernoulli distribution of normal distribution", it is as if you were talking about "red color of blue color". Moreover, I can't see how you could sample "heads" from normal distribution (how -5.22454666 could be a "heads"?). Even more, even if somehow you sampled the "heads" from something, then tails are everything else, so they cannot have different distribution then heads! Your question does not make any sense at all. – Tim Jan 18 '18 at 15:37
  • Despite the incorrect terminology, I find the question perfectly clear @Tim: it describes a mixture of two Normal distributions. Perhaps if you were to read "mixture" in place of "bernoulli" every place it occurs, the meaning would be more apparent. – whuber Jan 18 '18 at 15:43
  • @whuber maybe you're right, the wording completely lost me. – Tim Jan 18 '18 at 15:44

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