I have a probability distribution $f(x) = c\cdot\exp(-c\cdot x)$ where $c\in\Bbb R$ and $ x \in [0, \infty ) $
I have to make a collection of 5 randomly selected values from this distribution. $ x_1,x_2,x_3,x_4, x_5 $ which are randomly selected based on the probability distribution. (I would expect more frequent values at small $x$ because its more probable.. exponential decay of probability).
Now let's say $ n = x_1 + x_2 + x_3+ x_4 + x_5 $
What would be the new probability distribution for the collection wrt 'n' i.e. p(n)?
I am not sure how to start. I understand that I could use multiplication of probabilities for a 'n': $ f(x_1)\cdot f(x_2)\cdots f(x_5) $ but how to include condition of $ x_1+x_2+x_3+x_4+x_5=n $ to find p$(n)$?
We have independence of the samples, and that they are identically exponentially distributed with rate parameter $c$.
Apply the Law of Total Probability.
When five non-negative values sum to a non-negative total ($n$), then four have to sum to at most $n$, and the fifth shall be the remainder.
$$\begin{align} p(n) &= \mathbf 1_{0\leqslant n}\cdot{\mathop{\iint\!\!\!\iint}\limits_{s+t+u+v\leq n} f(s)f(t)f(u)f(v)f(n-s-t-u-v)\,\mathrm d v\,\mathrm d u\,\mathrm d t\,\mathrm d s}\\&=\mathbf 1_{0\leq n}\cdot\int_0^n\int_0^{n-s}\int_0^{n-s-t}\int_0^{n-s-t-u}c^5\mathrm e^{-c n}\,\mathrm d v\,\mathrm d u\,\mathrm d t\,\mathrm d s\\&~~\vdots\end{align}$$