Exact distribution of sample mean of a continuous distribution Please first note that I do know about the central limit theorem but I wish to derive an exact expression for the sample mean for any continuous distribution with probability density function $f(x)$. I thought I derived it correctly here but there is some mistake I can't spot. Can anyone tell me where I have made a mistake?
I take a simple case to derive the sample mean for with only 2 samples. The probability density function $f(x)$ is defined as:
$$f(x)=\frac{x}{50}$$
The maximum value of $x$ is $10$ and the minimum value of $x$ is $0$. Note that:
$$\int_0^{10} f(x) dx =1$$
The samples are random variables $X_1$ and $X_2$
$$\bar{x}=\frac{x_1+x_2}{2}$$
If the sample mean is to equal $\bar{x}$ then given that the first sample is $x_1$ the second sample $x_2$ must equal $2\bar{x}-x_1$. But if $x_1$ is too large or too small then a value of $x_2$ may not exist in the range [0,10].
We know
$$x_2 \leq 10 \quad  \therefore 2\bar{x}-x_1 \leq 10$$
$$\implies x_1 \geq 2\bar{x}-10$$
And
$$x_2 \geq 0 \quad \therefore x_1 \leq 2\bar{x}$$
$$\implies 2\bar{x}-10 \leq x_1 \leq 2\bar{x}$$
The probability that $X_1$ takes the value $x_1$ and $X_2$ takes the value $2\bar{x}-x_1$ is $f(x_1)f(2\bar{x}-x_1)$
To account for all cases of $X_1$ that could result in $\bar{X}=\bar{x}$ we integrate with respect to $x_1$
$$\int_{2\bar{x}-10}^{2\bar{x}}f(x_1)f(2\bar{x}-x_1)dx_1$$
I did this integration and it ended up as
$$\frac{1}{7500}(1000-570\bar{x}-120\bar{x}^2)$$
The sample mean must lie inside $[0,10]$ so the integral of all the eventualities of the sample mean over the range $[0,10]$ should equal 1. But:
$$\int_0^{10} \frac{1}{7500}(1000-570\bar{x}-120\bar{x}^2)=-7.8$$
So where have I made an incorrect assumption?
 A: This is rather trickier than it appears. The reason is that since the variables involved are non-negative, some subtle restrictions on the domain of integration of the convolution arise, resulting also in a piece-wise density function for $\bar X$.
I assume that $X_1$, $X_2$ are independent. We start by defining the variables $$Y_i = \frac12 X_i,\; i=1,2$$
Then by the change-of variables formula we easily obtain
$$f_{Y_i}(y_i) = f_{X_i}(2y_i)\cdot 2 = \frac{2}{5^2}y_i,\; i=1,2\;, y_i \in [0,5]$$
Define now the r.v. $\bar X = Y_1 + Y_2$.
In general, the convolution here is 
$$f_{\bar X}(\bar x) = \int_{-\infty}^{\infty} f_{Y_1}(\bar x - y_2)f_{Y_2}(y_2)dy_2 $$
...where for some range of values the integral will be zero. But the question is, what will be the effective limits of integration? Will they be "just" $[0,5]$ which is the range of $y_2$, or something else?
If we look at the distribution function of $\bar X$ we have
$$F_{\bar X}(\bar x) = P(\bar X \le \bar x) = P(Y_1 + Y_2 \le \bar x) $$
This gives us the inequality $y_1 + y_2 \le \bar x \Rightarrow y_1\le \bar x - y_2$. BUT $y_1$ ranges in $[0,5]$ so we must respect 
$$0 \le y_1 \le \bar x - y_2 \le 5 \Rightarrow 0 \le \bar x - y_2 \le 5$$
which is analyzed in two inequalities, 
$$y_2 \le \bar x\;,\qquad \text{and} \;\bar x -5 \le y_2 $$
Now $y_2$ ranges in $[0,5]$, while $\bar x$ ranges in $[0,10]$. So we have to break the range of $\bar x$ accordingly.  
A) $0\le \bar x \le 5$.  
When $\bar x$ ranges in this interval, the inequality $y_2 \le \bar x$ is binding, while the inequality  $\bar x -5 \le y_2 $ is always satisfied (= for all values of $y_2$). So for this range of $\bar x$ the range of integration of the convolution will be $[0, \bar x]$.  
B)  $5\le \bar x \le 10$.
For this range the inequality $y_2 \le \bar x$ is always satisfied, while the inequality $\bar x -5 \le y_2 $ is binding. So for this range of $\bar x$ the range of integration of the convolution will be $[\bar x -5, 5]$.
Therefore we have,
$$\text{For}\; 0\le \bar x \le 5\;\;,  f_{\bar X}(\bar x) = \int_{0}^{\bar x} f_{Y_1}(\bar x - y_2)f_{Y_2}(y_2)dy_2 = \frac{4}{5^4}\int_{0}^{\bar x} (\bar x-y_2)y_2dy_2 $$
while 
$$\text{For}\; 5\le \bar x \le 10\;\;,  f_{\bar X}(\bar x) = \int_{\bar x -5}^{5} f_{Y_1}(\bar x - y_2)f_{Y_2}(y_2)dy_2 = \frac{4}{5^4}\int_{\bar x -5}^{5} (\bar x-y_2)y_2dy_2 $$
Carrying out the simple but tedious algebra we arrive at
$$0\le \bar x \le 5\qquad   f_{\bar X}(\bar x) = \frac {2}{3\cdot 5^4}\bar x^3 $$
$$5\le \bar x \le 10\qquad   f_{\bar X}(\bar x) = -\frac {8}{15} + \frac {4}{5^2}\bar x - \frac {2}{3\cdot 5^4}\bar x^3 $$
Not very nice-looking, eh? Still, it will be illuminating to draw the graph of the above density - it is very nice-looking. It is everywhere non-negative and has a nice curvature. Moreover, one can check that it integrates to unity, namely
$$\int_{0}^{10}f_{\bar X}(\bar x)d\bar x = \int_{0}^{5}\frac {2}{3\cdot 5^4}\bar x^3d\bar x  +\int_{5}^{10}\Big[-\frac {8}{15} + \frac {4}{5^2}\bar x - \frac {2}{3\cdot 5^4}\bar x^3\Big]d\bar x  $$
$$ = \frac 16 + \frac 56 = 1$$.
