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I have a continuous distribution function, but it has a mass point at the upper bound, say with a probability of p>0. How to calculate its variance?

Adam
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1 Answers1

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If $b$ is the upper bound then the variance is $\int x^{2} f(x)\, dx-(\int f(x) \, dx)^{2}+b^{2}p-bp$ where $f$ is the density up to $b$.

Kavi Rama Murthy
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  • Thanks, do you mind providing a derivation? – Adam Sep 16 '18 at 23:57
  • When you calculate the second moment you integrate over ${b}$ and then over the remaining part. This gives $b^{2}p+\int x^{2}f(x) , dx$. Similarly you can write down the first moment and then calculate the variance. – Kavi Rama Murthy Sep 17 '18 at 00:29
  • No, I don't think this is right. Something is wrong in your formula. – Adam Sep 18 '18 at 20:09