### Distribution is not a sufficient measure The expectation is not a sufficient measure to describe a distribution. It is possible for two distributions of completely different nature to have exactly the same expectation. #### Example Let’s look at the following two distributions: **Distribution A:** – Value: 2, Probability: 0.30
– Value: 3, Probability: 0.25
– Value: 5, Probability: 0.25
– Value: 7, Probability: 0.20
– Total: 1.00 **B distribution:** – Value: 0, Probability: 0.25
– Value: 1, Probability: 0.25
– Value: 7, Probability: 0.25
– Value: 8, Probability: 0.25
– Total: 1.00 Note: The above distributions are not the results of any known realistic experiment, but we can assume that such an experiment exists. #### Calculating the Expectation – Distribution A: `4=7*0.2+5*0.25+3*0.25+2*0.3`
– Distribution B: `4=8*0.25+7*0.25+1*0.25+0*0.25` The mean of both distributions is the same, 4. #### Are the two distributions the same? It is easy to see that distribution B is more spread out around the mean than distribution A. Therefore, we will use the measure of dispersion. #### Calculating the standard deviation **Distribution A:** – Value: 2, Probability: 0.30, Deviation: `(-2)^2=4`
– Value: 3, Probability: 0.25, Deviation: `(-1)^2=1`
– Value: 5, Probability: 0.25, Deviation: `1^2=1`
– Value: 7, Probability: 0.20, Deviation: `3^2=9` Expected deviation squared: `3.5=9*0.2+1*0.25+1*0.25+4*0.3` Standard deviation: `1.87=sqrt(3.5)` **B’ distribution:** – Value: 0, Probability: 0.25, Deviation: `(-4)^2=16`
– Value: 1, Probability: 0.25, Deviation: `(-3)^2=9`
– Value: 7, Probability: 0.25, Deviation: `3^2=9`
– Value: 8, Probability: 0.25, Deviation: `4^2=16` Expected deviation squared: `12.5=16*0.25+9*0.25+9*0.25+16*0.25` Standard deviation: `3.54=sqrt(12.5)` As we expected, the standard deviation of distribution B is greater than the standard deviation of distribution A. Distribution B is more spread out than distribution A.
