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May 09'23
Exercise
Random variables [math]X[/math] and [math]Y[/math] have joint distribution
| X=0 | X=1 | X=2 | |
|---|---|---|---|
| Y=0 | 1/15 | a | 2/15 |
| Y=1 | a | b | a |
| Y=2 | 2/15 | a | 1/15 |
Let [math]a[/math] be the value that minimizes the variance of [math]X.[/math]
Calculate the variance of [math]Y[/math].
- 2/5
- 8/15
- 16/25
- 2/3
- 7/10
Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.
May 09'23
Solution: A
The marginal distribution of X has probability 1/5 + a at 0, 2a + b at 1, and 1/5 + b at 2. Due to symmetry, the mean is 1 and so the variance is (0 − 1) 2 (1/ 5 + a ) + (1 − 0) 2 (1/ 5 + a )= 2 / 5 + 2a which is minimized at a = 0. The marginal distribution of Y is the same as that of X and thus has the same variance, 2/5 + 0 = 2/5.
Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.