Suppose that $X_1,X_2,…,X_{10}$ are iid exponential random variables. What is the joint density of $X_1,…,X_{10}$ at $x_1,…,x_{10}?$ Is it the product $$f_{(X_1,\dots,X_N )}(x_1, \dots , x_n) = f_{X_1}(x_1)f_{X_2}(x_2)\dots f_{X_n}(x_n)$$ or is it different for an exponential?
Joint density of Independent and identically distributed Exponential random variables
density functionexponential distributionprobabilityprobability distributionsrandom variables
Related Solutions
If I were you, I would use transformations.
$$ Y1 = X1 - X2 = u_1(x_1,x_2)$$ $$ Y2 = X1 + X2 = u_2(x_1,x_2)$$ $$ X_1 = \frac{Y_1+Y_2}{2}=w_1(y_1,y_2)$$ $$ X_2 = \frac{Y_1-Y_2}{2}=w_2(y_1,y_2)$$
$$ f(y_1,y_2) = f(w_1(y1, y2),w_2(y1,y2)|J|$$
j is jaconian
I solved same problem with standard normal distributions. I added picture of my process. or You can solve this problem by using distribution method.
good luck :)
Hint: Use the transformation theorem.
If $Y=(Y_1,Y_2),$ where $Y_1$ and $Y_2$ are as described by you, $Y=g(X)$, $X=(X_1,X_2)$, then:-
$f_Y(y)=|J(y)|f_X(g^{-1}(y))$, if $y$ is on g's range (and zero otherwise).
Note that $J$ is the jacobian of the inverse transformation, $g^{-1}(Y)$.
Best Answer
For any independent random variables $X_1, \dots, X_n$ with respective probability density functions $f_1(x), \dots, f_n(x)$ their joint probability density function is $f(x_1, \dots, x_n) = \prod_{i=1}^n f_i(x_i)$. Note that $X_i$ do not need to be exponential and do not need to have the same distribution. Instead, the form of the joint distribution is the defining property of independence.