# Does convolution of a probability distribution with itself converge to its mean?

Suppose we have a probability density function $$f(x)$$ 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 mean of $$f(x)$$?

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...

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...

Here is the definition of convolution operation: https://en.wikipedia.org/wiki/Convolution_of_probability_distributions

• (i) What do you mean by "a probability distribution $f(x)$"? A probability distribution is a measure. (ii) What do you mean by $\lambda f$? If $f$ is indeed a probability distribution and $\lambda f$ is the usual product, then the convolutions will converge to the zero measure (even in total variation). (iii) How can "the resulting distribution converge[] to the mean"? Again, a probability distribution is a measure, whereas the mean is number. Commented Feb 10, 2022 at 16:34
• @IosifPinelis (iii) I think by "converging to the mean", she means that if we repeat the convolution operation, then the distribution will finally become a Dirac-delta distribution at the mean. (ii) Convolution means this" en.wikipedia.org/wiki/Convolution_of_probability_distributions. (i) $f$ usually means the probability density function of a distribution, I think.
– dodo
Commented Feb 10, 2022 at 17:08
• If $f$ is a probability density function then the convolution $f*f$ is also a probability density function, so if $0<\lambda<1$ is a scalar, then then $\lambda^2 f*f$ and $(1-\lambda)^2 f*f$ and $\lambda(1-\lambda)f*f$ are functions whose integrals are less than $1.$ and $f*f$ is the density function of a distribution whose variance is twice that of the distribution whose density is $f.$ So this is moving away from the delta function, not toward it. Commented Feb 10, 2022 at 17:24
• I wonder whether what you want is a weighted average of i.i.d. random variables, rather than assigning weights to the density functions? Commented Feb 10, 2022 at 17:54
• @MichaelHardy it is the weighted average of i.i.d random variables. The answer provided below is a more accurate description of my problem and I agree with his answer.
– Emma
Commented Feb 14, 2022 at 9:20

$$\newcommand{\si}{\sigma}\newcommand{\ep}{\varepsilon}\newcommand\E{\operatorname{\mathsf{E}}}\newcommand\Var{\operatorname{\mathsf{Var}}}\newcommand{\de}{\delta}$$I think a meaning can be attached to your post as follows: You appear to confuse three related but quite different notions: (i) a random variable (r.v.), (ii) its distribution, and (iii) its pdf. (Unfortunately, many people do so.)

So, my guess at what you were trying to say is as follows: Let $$X$$ be a r.v. with values in $$[a,b]$$. Let $$\mu:=\E X$$ and $$\si^2:=\Var X$$. Let $$X_{\cdot,\cdots}$$ with various indices $$_{\cdot,\cdots}$$ denote independent copies of $$X$$. Let $$t:=\lambda\in(0,1)$$.

At the first step, we take any $$X_1$$ and $$X_2$$ (which are, according to the above convention, two independent copies of $$X$$). We multiply the r.v.'s $$X_1$$ and $$X_2$$ (not their distributions or pdf's) by $$t$$ and $$1-t$$, respectively, to get the independent r.v.'s $$tX_1$$ and $$(1-t)X_2$$. The latter r.v.'s are added, to get the r.v. $$$$S_1:=tX_1+(1-t)X_2,$$$$ whose distribution is the convolution of the distributions of the r.v.'s $$tX_1$$ and $$(1-t)X_2$$.

At the second step, take any two independent copies of $$S_1$$, multiply them by $$t$$ and $$1-t$$, respectively, and add the latter two r.v.'s, to get a r.v. equal in distribution to the r.v. $$$$S_2:=t(tX_1+(1-t)X_2)+(1-t)(tX_3+(1-t)X_4) \\ =t^2 X_1+t(1-t)X_2+(1-t)tX_3+(1-t)^2X_4;$$$$ (in accordance with the indexing convention, the r.v.'s $$X_1,X_2,X_3,X_4$$ are independent copies of $$X$$).

Continuing so, after the $$n$$th step we get a r.v. $$$$S_n:=\sum_{k=0}^n t^k(1-t)^{n-k}\sum_{i=1}^{\binom nk} X_{n,k,i}$$$$ (in accordance with the indexing convention, the $$X_{n,k,i}$$'s are independent copies of $$X$$). This expression for $$S_n$$ can be checked by induction on $$n$$. (One can also use/check the following re-indexed expression for $$S_n$$: $$$$\sum_{(\de_1,\dots,\de_n)\in\{0,1\}^n}X_{\de_1,\dots,\de_n} \prod_{j=1}^n(t^{\de_j}(1-t)^{1-\de_j});$$$$ imagine the corresponding binary tree.)

Now we have \begin{aligned} \E S_n&=\sum_{k=0}^n t^k(1-t)^{n-k}\sum_{i=1}^{\binom nk} \E X_{n,k,i} \\ &=\sum_{k=0}^n t^k(1-t)^{n-k}\binom nk \mu =\mu \end{aligned} and \begin{aligned} \Var S_n&=\sum_{k=0}^n (t^k(1-t)^{n-k})^2 \sum_{i=1}^{\binom nk} \Var X_{n,k,i} \\ &=\sum_{k=0}^n (t^2)^k((1-t)^2)^{n-k}\binom nk \si^2 \\ &=q^n\si^2\to0 \end{aligned}
(as $$n\to\infty$$), where $$q:=t^2+(1-t)^2\in(0,1)$$.

So, by Chebyshev's inequality, for any real $$\ep>0$$, $$$$P(|S_n-\mu|>\ep)\le\frac{\Var S_n}{\ep^2}=\frac{q^n\si^2}{\ep^2}\to0.$$$$

Thus, indeed $$S_n\to\mu$$ in probability.

Moreover, since $$$$\sum_{n=1}^\infty P(|S_n-\mu|>\ep)\le\sum_{n=1}^\infty\frac{q^n\si^2}{\ep^2}<\infty$$$$ for any real $$\ep>0$$, it follows by the Borel–Cantelli lemma that $$S_n\to\mu$$ almost surely.

Below is the image of a Mathematica notebook illustrating the geometrically fast convergence of $$S_n$$ to the mean $$\mu$$. It takes Mathematica about 0.01 sec to simulate values of $$S_1,\dots,S_{10}$$ and about 9 sec to simulate values of $$S_1,\dots,S_{20}$$ (in the second case, we deal with a binary tree with $$2^{20}\approx10^6$$ leaves).