Sept 2011 Q.7 (ii)

[madhurikumar]

Type 1 risk - 80% prob - Normal
Type 2 Risk - 20% prob - Normal

While I attempted this question , I presumed the loss distribution of the sum of these two Normal distributions was Normal.

Could someone throw more light on this part of the solution and explain why the cumulative pdf isn't normal ?

Thanks in advance.

 

[suraj]

Type 1 risk - 80% prob - Normal
Type 2 Risk - 20% prob - Normal

While I attempted this question , I presumed the loss distribution of the sum of these two Normal distributions was Normal.

Could someone throw more light on this part of the solution and explain why the cumulative pdf isn't normal ?

Thanks in advance.

X is not sum of two Normal variables. It's a mixture of two Normal RV X1 and X2.

So

\(f_X(x) = 0.8~f_{X_1}(x_1) + 0.2~f_{X_2}(x_2) \)

not X = 0.8 X1 + 0.2 X2

See previous question on this paper which is also on Mixture Distribution.

 

[madhurikumar]

Argh. Yes I get it. Thank you Suraj.

 

[Jammy]

Hi, I still have the same doubt as madhuri, can someone please elaborate and clear this up?

 

[Katherine Young]

Suraj is perfectly right:

We don't have \(X = 0.8 X1 + 0.2 X2\) (which would be Normal).

Instead we have \(fx(x)=0.8fx_1(x_1)+0.2fx_2(x_2)\) (which isn't Normal).

To check this isn't Normal, you can substitute in the PDFs of a normal distribution into this equation, (one PDF for \(fx_1(x_1)\) and another PDF for \(fx_2(x_2)\), then try to rearrange your equation to make the whole thing look like just one PDF of a Normal distribution. You won't be able to do it.

Try it and see!

If you still don't believe us, google it! eg https://en.wikipedia.org/wiki/Mixture_distribution