# Translation from my Japanese blog (2)

Written by actuary on Thu Oct 09, 2008 2:24 am in blog actuary_math's blog under Actuary_math's blog -
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In the last post I left proof of a proposition.

So I will prove it as below.

We must use the Lebesgue's integral theory to prove the proposition.

Proposition

Suppose that X and Y are independent random variables with probable density functions f(x) and g(y) respectively.

And suppose that f(x) and g(y) are continuous almost everywhere, and that a least one of f(x) and g(y) is bounded.

Then h(s),the p.d.f. of S=X+Y is continuous at ANY real number s.

Proof

As

h(s)=\int f(s-y)g(y)dy=\int f(x)g(s-x)dx

(In the above formula "\int" means integral from -infinity to +infinity.)

it is sufficient that we suppose that f(x) is bounded.

That is there is a positive real number K that f(x)<K for all x.

And it is sufficient for us to prove that for any sequence {s_n} that converges to s

\int f(s_n-y)g(y)dy =h(s)

Note that

f(s_n-y)g(y) <= K*g(y)

and that

\int K*g(y)=K*1=K<+infinity

(because h(s)=\int g(y)dy=1)

On the other hand

for almost any y

f(s_n-y)g(y) converge to f(s-y)g(y).

(because f(x) and g(y) are continuous almost everywhere)

So we get

\int f(s_n-y)g(y)dy =h(s)

by dominated convergence theorem(or Lebesgue's convergence theorem)

q.e.d.

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