excans:5df2b78412: Difference between revisions
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(Created page with "'''Solution: C''' <math display = "block"> \operatorname{E}(\frac{X}{1-X}) = 60 \int_0^1 \frac{x}{1-x} x^3 (1-x)^2 dx = 60 \int_0^1 x^4(1-x) dx = 60(x^5/5 - x^6/6) \Big |_0^1...") |
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E[(\frac{X}{1-X})^2] = 60 \int_0^1 \frac{x^2}{(1-x)^2} x^3 (1-x)^2 dx = 60 \int_0^1 x^5 dx = 60(x^6/6 ) \Big |_0^1 = 60(1/6) = 10 | \operatorname{E}[(\frac{X}{1-X})^2] = 60 \int_0^1 \frac{x^2}{(1-x)^2} x^3 (1-x)^2 dx = 60 \int_0^1 x^5 dx = 60(x^6/6 ) \Big |_0^1 = 60(1/6) = 10 | ||
</math> | |||
<math display = "block"> | |||
\operatorname{Var}\left( \frac{X}{1-X}\right) = 10-2^2 = 6. | |||
</math> | </math> | ||
{{soacopyright | 2023}} | {{soacopyright | 2023}} | ||
Latest revision as of 00:32, 8 May 2023
Solution: C
[[math]]
\operatorname{E}(\frac{X}{1-X}) = 60 \int_0^1 \frac{x}{1-x} x^3 (1-x)^2 dx = 60 \int_0^1 x^4(1-x) dx = 60(x^5/5 - x^6/6) \Big |_0^1 = 60(1/5 -1/6) = 2
[[/math]]
[[math]]
\operatorname{E}[(\frac{X}{1-X})^2] = 60 \int_0^1 \frac{x^2}{(1-x)^2} x^3 (1-x)^2 dx = 60 \int_0^1 x^5 dx = 60(x^6/6 ) \Big |_0^1 = 60(1/6) = 10
[[/math]]
[[math]]
\operatorname{Var}\left( \frac{X}{1-X}\right) = 10-2^2 = 6.
[[/math]]
Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.