Miller Ch4 Q5 Pg 76 Study Notes

[INDRAJIT]

Hi David,

Didn't really understand the approach to this problem.

A random variable X has
a density function that is a normal mixture with two independent
components: the first normal component has an expectation (mean) of 4.0 with variance of 16.0;
the second normal component has an expectation (mean) of 6.0 with variance of 9.0. The
probabil
ity weight on the first component is 0.30 such that the weight on the second component
is 0.70. What is the probability that X is less than zero; i.e., Prob [X<0]?
a)
0.015%
b)
1.333%
c)
6.352%
d)
12.487%

My understanding was that we calculate the mean of the mixture = 0.3*4 + 0.7*6 = 5.4
Similarly, variance = 0.3*16 + .7*9 = 11.1
z = (0-5.4)/((11.1)^0.5) = 1.6208
Now corresponding to z(1.62) is the area 0.4474 from the mean. Hence, P(z<1.62) = 0.5-0.4474 = 5.26%

What is wrong with this approach? The answer with the different approach given in the notes is 6.352%.

 

[David Harper CFA FRM]

Hi @INDRAJIT

Source question is here, fwiw, at https://forum.bionicturtle.com/threads/p1-t2-312-mixture-distributions.7103/

There are two problems with your approach:

1. The variance is not quite correct, the variance of this mixture distribution is a bit more complicated (I get 11.94) but it's not required for the question (and is out of scope)
2. More importantly, your step "z = (0-5.4)/((11.1)^0.5) = 1.6208" assumes the mixture is a normal distribution. That's instructive (and totally understandable)! Because it highlights the difference between adding random normal variables (which does create a normal! i.e., the summation stability property of the normal distribution says that when we add normals we get another normal). However, a mixture of two normals (a "normal mixture distribution") is different than adding two normals: the mixture distribution is more like the "piecewise construction" of a new density function based on components, please see http://en.wikipedia.org/wiki/Mixture_distribution. Anytime we employ Z values, we might ask "why I am assuming normality here?" I hope that helps,