does convolution of a probability distribution with itself converge to its mean Suppose we have a probability distribution $f(x)$ with a finite support $[a,b]$. If we take the probability convolution of $\lambda f $ with $(1-\lambda)f,0 <\lambda<1$ recursively for many times, does the resulting distribution converges to the Dirac-delta distribution at the mean of $X$?
To be more specific: suppose $f_1(x)$ is probability distribution resulting form the convolution of $\lambda f $ with $(1-\lambda)f$, the second convolution would be $\lambda f_1 $ with $(1-\lambda)f_1$ and so forth...
Alternatively this can be explained in terms of random variables. First we use $'$ to denote the independent copy of a random variable, so $X'$ is an independent copy of the random variable $X$. Let $Y_0=\lambda X+(1-\lambda)X'$, $Y_1=\lambda Y_0+(1-\lambda)Y_0'$ ...,$Y_n= \lambda Y_{n-1}+(1-\lambda)Y'_{n-1}... $. Does $Y_n$ converge to a Dirac-delta distribution at the mean of $X$?
Could someone help with a formal proof? I tried to run some simulation with a discrete probability distribution with three outcomes. It seems it would converge as I increase the converge times from 1 to 3 times . But trying $4$ times crashes my laptop...
 A: $Y_0=\lambda X + (1-\lambda) X'$ so
$\text{var}(Y_0) = (\lambda^2 + (1-\lambda)^2)\text{var}(X)$
define $v = (\lambda^2 + (1-\lambda)^2)$
note $0.5< v < 1$ for $0< \lambda<1$
$Y_{1}=\lambda Y_0 + (1-\lambda) Y_0'$ and we have the general pattern
$Y_{n+1}=\lambda Y_n + (1-\lambda) Y_n'$
since $Y_n$ and $Y_n'$ are independent copies
$\text{var}(Y_{n+1})= v Y_n$
$Y_{n+1}=\lambda Y_n + (1-\lambda) Y_n'$ so
$\text{var}(Y_{n}) = v^{n+1}\text{var}(X)$ so variance goes to zero and  $\text{mean}(Y_n)=\text{mean}(X)$
A: This is just an example that your statement seems to be wrong, at least in some scenarios.
Suppose there are two same but independent distributions:
$Y_0$: 50% we get 1, 50% we get 0.
$Y_0'$: 50% we get 1, 50% we get 0.
Let $Y_1=1/2Y_0+1/2Y_0'$
Then $Y_1$ is: 25% we get 1, 50% we get 0.5, 25% we get 0.
The 25% comes from $50\%\times50\%$.
Then we have $Y_1$' totally same as $Y_1$. $Y_1'$ and $Y_1$ are independent.
By repeating, it seems that the probability of $Y_n=0.5$ can never exceed $50\%$
