What is the expected value of this random number?

Question

Suppose $a$ is a random number between 0 and 1 and suppose $b$ is a random number $\in (0,1)$ also.

What is the expected value of $a^b$? (i.e. If I performed this operation $n$ times, what would be the average value obtained as $n \rightarrow \infty$?

Answer 1

Fix $b$, then the average value of $a^b$ is $\int_0^1 a^b da=1/(b+1)$.
Then the average value of $1/(b+1)$ is $\ln2$.
On the other hand, as Jano says in comments, I don't know whether the order of integration matters.

Answer 2

The expected value is $\int_Q a^b dA$, where $Q$ is the unit square $[0,1]\times[0,1]$ (define $0^0=1$ if you must).

Since everything is well-behaved, $\int_Q a^b dA=\int_0^1\int_0^1 a^b \,db \,da=\int_0^1\int_0^1 a^b \,da\,db$.

You can now proceed as in Michael's answer.

Answer 3

Both answers of lhf and Michael are, of course, correct. Though, to complete them, I looked into the limit case for which you ask in your question, that is for a sequence of independent $U_1, U_2, \dots, U_n$ uniformly distributed over $(0,1)$, what is the expectation of $(((U_1^{U_2})^{U_3})\dots)^{U_n} = U_1^{{U_2}{U_3}\dots{U_n}}$?

We know that for each $U_i$ is the variable $-\log U_i$ exponentially distributed with parameter $1$. Therefore, for $n>1$ the distribution of $Y = -\log \prod_{i=2}^n U_i = \sum_{i=2}^{n} - \log U_i$ is gamma with parameters $k=n-1$ and $\theta = 1$, i.e. having density $$ g(y) = \frac{y^{n-2} \exp(-y)}{\Gamma(n-1)} \mbox{ for }y>0, $$ and $Y$ is independent of $U_1$. Moreover, $\prod_{i=2}^n U_i = \exp(-Y)$. Therefore, we can directly compute the expected value by $$ \mathbf E U_1^{U_2 \dots U_n} = \mathbf E U_1^{\exp(-Y)}= \int_0^1 \int_0^\infty u_1^{\exp(-y)} g(y) d y d u_1, $$ and after some computation we end up with $$ \mathbf E U_1^{U_2 \dots U_n} = (2^{2-n}-1)\zeta(n-1) \mbox{ for }n>2, $$ where $\zeta$ is the Riemann zeta function. Finally, as $n\to\infty$ the expectation converges to $1$.