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I have a Beta distribution, and am interested in calculating expectations and conditional expectations. The domain on the distribution is $z \in [0,1]$ Ignoring constants of proportionality, the expectation of an exponential function is:

$Assumptions = z >= 0 && z <= 1 && a > 0 && b > 0 && c > 0 && d > 0 && d < 1;

Print["*******Different parameters, {a,b}"];
Integrate[Exp[c z]  z^(a - 1) (1 - z)^(b - 1), {z, 0, 1}]

Print["********Symmetric parameter, a"];
Integrate[Exp[c z]  z^(a - 1) (1 - z)^(b - 1) /. {b -> a}, {z, 0, 1}]

The output from this is:

Gamma[a] Gamma[b] Hypergeometric1F1Regularized[a, a + b, c]
(-c)^(1/2 - a) E^(c/2) Sqrt[\[Pi]]  BesselI[-(1/2) + a, -(c/2)] Gamma[a]

These are solid. The problem I am having is when calculating conditional expectations, which is effectively integration over a subset of the domain. Again ignoring constants of proportionality,

Print["********Truncated Integral on [d, 1], symmetric for simplicity"];
Integrate[Exp[c z]  z^(a - 1) (1 - z)^(b - 1) /. {b -> a}, {z, d, 1}]

This is unable to solve it. Any idea on this? Are there any tricks in mathematica, in math that I am missing?

share|improve this question
    
{z, d, 1} are you sure it is appropriate ? –  Rorschach Oct 8 '13 at 17:25
    
for some fixed $0 \leq d < 1$, why wouldn't it be? –  jlperla Oct 8 '13 at 17:28
    
Integrate of a function is not from range of R to constant.It is from constant to constant or function to function. –  Rorschach Oct 8 '13 at 17:37
1  
I am very confused. Isn't the following legitimate for example? Integrate[z, {z, d, 1}] --> 1/2 - d^2/2 –  jlperla Oct 8 '13 at 17:43
    
you shall try integration by parts, it shall help.d in assumption doesn't play any role and is considered a symbol if you place it in range part. –  Rorschach Oct 8 '13 at 17:53

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