Independent solution

How to solve this Multinomial Distribution question

Answer in brief

The two category counts have expectations 0.4 and 0.2. Applying the two loss amounts linearly gives expected total loss 2(0.4)+20(0.2)=4.80, which matches choice E.

Setup

Setup

Let I and J count the two nonzero outcome categories among the two trials. Their joint distribution is multinomial, so each count has expectation equal to the trial count times its category probability.

E⁡[I]=2(0.20)=0.40\operatorname{E}[I]=2(0.20)=0.40
E⁡[J]=2(0.10)=0.20\operatorname{E}[J]=2(0.10)=0.20

Model

Model

Express the total loss as a linear combination of the category counts.

L=2I+20JL=2I+20J
E⁡[L]=2E⁡[I]+20E⁡[J]\operatorname{E}[L]=2\operatorname{E}[I]+20\operatorname{E}[J]

Compute

Compute

Substitute the two expected counts. No independence between I and J is needed for this expectation.

E⁡[L]=2(0.40)+20(0.20)\operatorname{E}[L]=2(0.40)+20(0.20)
E⁡[L]=0.80+4.00=4.80\operatorname{E}[L]=0.80+4.00=4.80

Answer

Answer

The expected total loss is 4.80.

E⁡[L]=4.80(E)\boxed{\operatorname{E}[L]=4.80\quad\text{(E)}}