We have $\xi_i \geq 0$, $\forall i = \overline{1,n}$ (exponetial i.i.d. variables). Assume that $S_n = \xi_1 +...+ \xi_n$.
It is easy to show that $\mathrm{E} (\xi_1\vert S_n = 1) = \frac{1}{n}$.
$$f_{\xi\vert\eta} = \frac{f_{\xi\eta}(x,y)}{f_{\eta}(y)}$$
On the other hand: $\mathrm{E} (\xi_1\vert S_n = 1) = \int\limits_0^1 \frac{xf^{*(n-1)}(1-x)}{f^{*n}(1)}\:\mathrm{d}x$, where $$f^{*n}(x) = \idotsint\limits_{z_{1}+...+z_{n} = x}f(z_1)...f(z_n)\:\mathrm{d}z_1\dots\:\mathrm{d}z_n;\quad f(t) = \lambda e^{-\lambda t}$$
Am I right? I suppose that I am not.
I think that in exponential case $f^{*n}(t) = \frac{\lambda e^{-\lambda t}(\lambda t)^{n-1}}{\Gamma(n)}$ because of the fact that sum of exponential i.i.d. variables has gamma distribution.
For example, let $n=2$.
$$\frac{1}{2} = \mathrm{E} (\xi_1\vert S_n = 1) = \int\limits_0^1 \frac{x\lambda e^{-\lambda (1-x)}}{\lambda^2 e^{-\lambda }}\:\mathrm{d}x \neq \frac{1}{2}$$
So tell me please where I'm wrong and how to calculate such expectations correctly.
P.S. This question has appeared after my other question where @Did commented an answer to an exponential case, but I can't calculate the result. Here is a link. Conditional expectation of second moment given sum of iid variables.
So, the error was in the formula for the conditional expectation. Correct formula: $$\mathrm{E} (\xi_1\vert S_n = 1) = \int\limits_0^1 \frac{xf(x)f^{*(n-1)}(1-x)}{f^{*n}(1)}\:\mathrm{d}x.$$