Convolution of probabilities

Question

It is a well known fact that for a random variable $Z=Y_1+Y_2+...+Y_n$ where $Y_i$ are independently distributed then the probability density function of $Z$ is the convolution of the density functions of the random variables $Y_i$ for $i=1,2,...,n$. In other words if $f_Z$ stands for the density function of $Z$ then $$f_Z=f_{Y_1}\ast f_{Y_2}\ast...\ast f_{Y_n}$$ where $\ast$ stands for the convolution operation. However is the converse true? If we have $f_Z=f_{Y_1}\ast f_{Y_2}\ast...\ast f_{Y_n}$ does it follow that $Z=Y_1+Y_2+...+Y_n$ where $Y_i$ are independently distributed. This might be an elementary fact but I can not see an immediate way to prove or disprove this. Any suggestion would be more than welcome.

Answer 1

Perhaps this is a counterexample to what you think may be true.

Let $U_1, U_2$ be iid $N(\mu=9, \sigma^2 = 9).$

Let $V_1, V_2, V_3$ be iid $N(6, 6).$

Let $W_1, W_2, \dots, W_9$ be iid $N(2, 2).$

Let $X_1, X_2, \dots, X_{18}$ be iid $N(1, 1).$

Then $\sum_{i=1}^2 U_i,\; \sum_{i=1}^3 V_i,\; \sum_{i=1}^9 W_i,\;$ and $\sum_{i=1}^{18} X_i$ are all distributed $N(18, 18).$

So if you know that $Z \sim N(18, 18),$ then there is no unique way to express it is a sum of iid random variables.

As a practical matter, because of the Central Limit Theorem, if you knew that a normal random variable is the sum of $n = 1000$ iid random variables, you wouldn't be able to guess whether the 1000 'pieces' were distributed normal, uniform, Poisson, or according any of a number of other distributions.

Your question is not exactly clear to me. If you had something else in mind, please leave a Comment and I (or someone else) will likely be able to help.