random variables with uniform probability and expectancy Suppose we choose a random number n from 0..100.Y and X are discrete random independent variables that have uniform distribution with 0...n.If Y+X equals to 15 what is expectancy of n?
 A: It is probably worth starting by working out the answer to: If $Y$ and $X$ are discrete random independent variables that have uniform distribution on the integers $0,\ldots...n$, then what is the probability that $Y+X=15$. 
You could then use these probabilities as likelihoods in your original question.
A: Following up Henry's answer and comment, if we assume than $n$ is a fixed but unknown number and $X$ and $Y$ are independent discrete random variables uniformly distributed on $\{0, 1, \ldots, n\}$, then 
$$P\{X + Y = i\} = \begin{cases} (i+1)/(n+1)^2, & 0 \leq i \leq n,\\
(2n+1-i)/(n+1)^2, & n+1 \leq i \leq 2n,\end{cases}$$ with a maximum value
of $(n+1)^{-1}$ at $i = n$.  Given $X + Y = 15$, we know that $n$ must be at least $8$.  Furthermore, the likelihood of the observation $X + Y = 15$ is $0$ for $0 \leq n < 8$,$2/9^2$ when $n = 8$, $4/10^2$ when $n = 9$, etc., peaking at $1/16$ when $n = 15$/  Thus, as Henry suspected, the maximum-likelihood estimate of $n$ is $15. But the question asked by the OP is
If $Y+X = 15$, what is expectancy of n?
which seems to imply that the OP wants the conditional expected value of $n$
given $X + Y = 15$, presumably under the  assumption that the prior 
distribution of $n$ is uniform on $\{0, 1, \ldots, 100\}$.  So now we need
to compute 
$$P\{n = m \mid X+Y = 15\} 
= \frac{\frac{1}{101}P\{X+Y = 15 \mid n = m\}}{
\frac{1}{101}\sum_{i = 0}^{100}P\{X+Y = 15 \mid n = i\} }$$
which presumably leads to the conditional median of $n$ being 32
and the conditional mean being over 39.5.  I wonder what the Bayesian
or maximum aposteriori probability estimate of $n$ is?
