Suppose $X \sim \Gamma[n_1,\lambda], Y \sim \Gamma[n_2,\lambda]$, and $X+Y \sim \Gamma[n_1 + n_2, \lambda]$
Can we say that $X$ and $Y$ are independent?
Here's what I think: Suppose $f,g$ are p.d.f.'s of $X,Y$, and $h$ be their joint p.d.f. It is basically saying
$$ \forall z>0,\int_0^z h(x,z-x)dx = \int_0^z f(x)g(z-x)dx $$
Does it infer $h(x,y) = f(x)g(y)$ a.e.?
Well, I know that if $$ \int_0^xf(u)du = \int_0^x g(u)du$$ By taking the derivative of $x$, I suppose it means $f = g$ a.e.
However, I don't know how to take the derivative of $z$ in the previous formula, since $x$ appears at the same time with $z$. If I take the derivative of $z$, since $z-z=0$, I'll get $h(z,0) = f(z)g(0)$, which doesn't seem to help.
We know that if $X$, $Y$ are independent, the additivity holds. Now I'm doing a problem that requires to prove the independence of two r.v.'s. I found that they follow Gamma distributions and they obey additivity, but I'm not sure if it is legitimate to say these two r.v.'s are independence.
Thanks!
In terms of characteristic functions, the question is whether variables are independent if the product of their characteristic functions is equal to characteristic function of their sum: $$\mathbb E[e^{itX}]\cdot \mathbb E[e^{itY}] = \mathbb E[e^{it(X+Y)}].\tag{1}$$ We know that the variables are independent iff for all $t,s\in \mathbb{R}$ $$\mathbb E[e^{itX}]\cdot \mathbb E[e^{isY}] = \mathbb E[e^{i(tX+sY)}],$$ which is much stronger than $(1)$. So it should be obvious that the answer is negative. But it is not so easy to construct a particular example.
I will construct an example in the case where $n_1 = n_2 = 1$, $\lambda = 1$, so that we have $\mathrm{Exp}(1)$ variables (if you'll understand this construction, then you'll be able to make a similar construction in arbitrary case).
Let us first reduce this problem to a discrete case, which is easier to handle. It is well known (and easy to prove) that for an exponentially distributed random variable $X$, its integer part $\lfloor X\rfloor$ and its fractional part $\{X\}$ are independent and the integer part has shifted geometrical distribution: $$ p_n := \mathbb{P}(\lfloor X\rfloor = n) = e^{-n}(1-e^{-1}), n\ge 0. $$
So we want to construct two dependent variables $X'$, $Y'$ such that $$\mathbb{P}(X' = n) = \mathbb{P}(Y' = n) = p_n, n\ge 0,\tag{2}$$ and $$\mathbb{P}(X'+Y' = n) = \sum_{k=0}^n p_kp_{n-k}, n\ge 0.\tag{3}$$ Start by setting $p_{k,j} = p_kp_j$. Now, for some very small $\epsilon>0$ distort six values: \begin{eqnarray} & & p'_{1,0}= p_{1,0}+\epsilon\quad & p'_{2,0}= p_{2,0}-\epsilon\\ & p'_{0,1}= p_{0,1}-\epsilon\quad & & p'_{2,1}= p_{2,1}+\epsilon \\ & p'_{0,2}= p_{0,2}+\epsilon\quad & p'_{1,2}= p_{1,2}-\epsilon & \end{eqnarray} and set $\mathbb{P}(X' = k,Y'=j) = p'_{k,j}$ for these six values and $\mathbb{P}(X' = k,Y'=j) = p_{k,j}$ otherwise. It is easy to see that such $X',Y'$ are dependent and satisfy $(2)$ and $(3)$.
Finally, let $X'',Y''$ be independent random variables, which are independent of $X'$, $Y'$ and have the same distribution as fractional part of $\mathrm{Exp}(1)$.
Setting $X = X'+X''$, $Y = Y'+Y''$, we arrive at the required example.