# Sum of independent Random Variables

## Homework Statement

Three yearly losses.
First: Exponential
Second & Third: Weibull

Losses are independent.
Find the 95% VaR of the min loss

## The Attempt at a Solution

My first thought was:
Let L be total loss, A be first Loss, B be second loss, C be third loss

Pr(L<x) = Pr(A<x) * Pr(B<x) * Pr(C<x)

But logically this seems wrong to me, because if Pr(A<x) = 100% then Pr(L<x) should be 100%.

So then I thought maybe:

Pr(L<x) = a_1 Pr(A<x) + a_2 Pr(B<x) + a_3 Pr(C<x)

but how do I choose a_1, a_2,a_3? Seems like these would change based on the values of A,B,C which makes them r.v. themselves!

The prof hinted that we just need to find the CDF and that to find the CDF we first should find the survival function keeping in mind the r.v. are independent.

I'm sort of out of ideas.

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

Three yearly losses.
First: Exponential
Second & Third: Weibull

Losses are independent.
Find the 95% VaR of the min loss

## The Attempt at a Solution

My first thought was:
Let L be total loss, A be first Loss, B be second loss, C be third loss

Pr(L<x) = Pr(A<x) * Pr(B<x) * Pr(C<x)

But logically this seems wrong to me, because if Pr(A<x) = 100% then Pr(L<x) should be 100%.

So then I thought maybe:

Pr(L<x) = a_1 Pr(A<x) + a_2 Pr(B<x) + a_3 Pr(C<x)

but how do I choose a_1, a_2,a_3? Seems like these would change based on the values of A,B,C which makes them r.v. themselves!

The prof hinted that we just need to find the CDF and that to find the CDF we first should find the survival function keeping in mind the r.v. are independent.

I'm sort of out of ideas.

You say that you think Pr(L<x) = a_1 Pr(A<x) + a_2 Pr(B<x) + a_3 Pr(C<x).

Why on earth would you ever think that? There are perfectly standard formulas for the probability distribution of a sum of independent random variables. These formulas are found in every probability textbook, or even in hundreds of on-line articles.

BTW: what you have written is not the standard cdf, which would be Pr(L <= x), with a non-strict inequality. What you wrote was long ago abandoned as the definition of a cdf.

Point taken. I'm going down this path now:
$F_{A+B+C} = P(A+B+C \le x)$
$= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{x-b-c} f_A(a) f_B(b) f_C(c) da db dc$

$= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} F_A(x-b-c) f_B(b) f_C(c) db dc$
Not sure where to go next...

Last edited:
Ray Vickson
Homework Helper
Dearly Missed
Point taken. I'm going down this path now:
$F_{A+B+C} = P(A+B+C \le x)$
$= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{x-b-c} f_A(a) f_B(b) f_C(c) da db dc$

$= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} F_A(x-b-c) f_B(b) f_C(c) db dc$
Not sure where to go next...

I doubt that the task is 'doable' analytically. Probably you need to resort to a numerical method.

Probably the easiest way is through successive convolution: if X = B+C, then its density is
$$f_X(x) = \int_0^x f_B(y) f_C(x-y) \, dy.$$
Then the density of S = A+B+C = A + X is
$$f_S(x) = \int_0^x f_A(y) f_X(x-y) \, dy$$
Note that the integrations only go from 0 to x, not from -∞ to +∞; this is because the random variables are all ≥ 0.

However, at this point I think you are stuck: you probably need to compute and store the values of ##f_X(x)## on a grid of x-values, or maybe come up with a convenient approximate formula for it, because in some cases at least we can prove that there is no finite formula for ##f_X## in terms of elementary functions. We can establish this by example: suppose B and C are iid Weibull with parameters k = 2 and λ = 1, so that the density of B (and C) is
$$f_B(x) = f_C(x) = 2x e^{-x^2}.$$
We can actually do the integral to get ##f_X##:
$$f_X(x) = x e^{-x^2} + \sqrt{\frac{\pi}{2}}\,(x^2-1)\, e^{ -x^2 /2}\, \text{erf}(x/\sqrt{2}).$$
It is impossible to give a finite, closed-form formula for this in terms of elementary functions, because if we could do it, we would have a finite, closed-form expression for 'erf', and that has been shown to be impossible. Of course, you can give non-finite expressions, such as infinite series and the like.

Anyway, that special case proves to be un-doable in simple terms, so the general case must also be un-doable.

All I can suggest is some type of numerical approximation when actual numerical inputs are specified. However, quantities like the mean and variance of S = A+B+C are easily obtained using standard result about moments of sums of independent random variables.