0
$\begingroup$

I have the following question :

$A,B,C$ are independent, $P(A)=0.2,P(B)=0.5,P(C)=0.8$, $N$ is a random variable that counts the number of incidents of $A,B,C$ that happened calculate Var(N)

I used indicators to calculate it yet, for some reason I get that $Var(N)<0$ which is wrong.

This is what I did:

$$Var(N)=E(N^2)-E(N)^2$$

$$E(N)=E(I_A)+E(I_B)+E(I_C)=0.2+0.5+0.8=1.5 \implies E(N)^2=2.25$$

$$E(N^2)=E(I_A^2)+E(I_B^2)+E(I_C^2)+E(I_A*I_B)+E(I_A*I_C)+E(I_B*I_C)$$

We know that by $E(N)$ that : $E(I_A^2)+E(I_B^2)+E(I_C^2)=E(I_A)+E(I_B)+E(I_C)=0.2+0.5+0.8=1.5$

since independent $$E(I_A*I_B)+E(I_A*I_C)+E(I_B*I_C)=0.2*0.5+0.2*0.8+0.8*0.5=0.66$$

Therefore we get that : $E(N^2)=1.5+0.66=2.16$

$$Var(N)=E(N^2)-E(N)^2=2.16-2.25=-0.09???$$

Any ideas what I did wrong I can't figure it out.

Thanks in advance :)

  • 1
    Shouldn't there be a factor of 2 in the squared term? This would give 1.5+1.32=2.82. – 2017-02-04
  • 0
    Hint: If you consider three Bernoulli random variables (equal 1 if a corresponding incident happens) then they will be independent. And your variable of interest is a sum of three. – 2017-02-04
  • 0
    @gobucksmath on which squared term are you talking about? – 2017-02-04
  • 0
    When you square $N$, there should be nine terms. You only have six terms. – 2017-02-04
  • 0
    @gobucksmath I still don't see it are you speaking about this term? which three term are lefted out? $E(N^2)=E(I_A^2)+E(I_B^2)+E(I_C^2)+E(I_A*I_B)+E(I_A*I_C)+E(I_B*I_C)$ – 2017-02-04
  • 1
    @JaVaPG it should be $E(N^2)=E(I_A^2)+E(I_B^2)+E(I_C^2)+2E(I_A*I_B)+2E(I_A*I_C)+2E(I_B*I_C)$ – 2017-02-04

0 Answers 0