# Mean of a truncated non-standard beta distribution

I have a non-standard beta distribution in the interval [-0.02 , 0.005] (as opposed to [0,1]).

I know its mean and variance (and thus α and β).

I want to calculate the mean of its truncation to [0 , 0.005].

See the following graph for clarification:

Is it possible to derive an equation for this problem?

Thank you.

There is a special function called the incomplete Beta function, $$B(x;a,b)=\int_0^x t^{a-1}(1-t)^{b-1}\text{d}t\qquad a,b>0$$ which serves both to normalise the density of a Beta $$\mathcal{B}(a,b)$$ distribution truncated to $$(c,d)$$: $$f(x;a,b,c,d)=\dfrac{x^{a-1}(1-x)^{b-1}}{B(d;a,b)-B(c,d)}$$ and to define its mean: $$\mathbb{E}_{a,b,c,d}[X]=\dfrac{B(d;a+1,b)-B(c;a+1,b)}{B(d;a,b)-B(c,d)}$$

• I see the Incomplete Beta Function requires a,b > 0 . My interval though is [-2%,0.5%]. Am I missing something? – Hugo González González Feb 6 '19 at 13:22
• I do not understand what an interval [-2%,0.5%] stands for. – Xi'an Feb 6 '19 at 13:27
• Please excuse my lack of clarity. If you look at the X axis of my second graph, that is the interval my Beta distribution is defined around. Then it is truncated at 0. I still call it a Beta distribution since i) it has the exact same shape as the one above, and ii) I understand generalized Beta distributions work with negative intervals – Hugo González González Feb 6 '19 at 13:56
• I will reword the question for further clarification. – Hugo González González Feb 6 '19 at 15:53
• You have to defined which Beta distribution you want to truncate to $[-0.02,0.005]$ and it cannot be a standard Beta since those have support $(0,1)$. If you truncate a distribution, the support of this distribution must be larger than the truncation region. – Xi'an Feb 6 '19 at 15:56