Question 11.4: The probability of an individual being hospitalized during t......
The probability of an individual being hospitalized during the year is P(H) = 0.15 = 15%, and once hospitalized the charges X have a continuous probability density function f_{X} \left(x\,|\,H=1\right)=0.1e^{-0.1x} for x > 0.
Determine the expected value, standard deviation and the ratio of the standard deviation to the mean of hospital charges.
The expected value of hospital charges is given by
E[X]~=~P\left({ H}\neq1\right)E\left[X\mid H\neq1\right]+P\left(H=1\right)E\left[X\mid H=1\right]= 0.85\mathrm{~*~0}+0.15\int_{0}^{\infty}x0.1e^{-0.1x}d x
= 0.15\int_{0}^{\infty}0.1x\ e^{-0.1x}d x
= 0.15(-x e^{-0.1x}|_{0}^{\infty}\;+\;\textstyle{\int_{0}^{\infty}}e^{-0.1x}d x)
= 0.15(0-10e^{-0.1x}|_{0}^{\infty})
= 1.5
Similarly, we may calculate E[X2] and then compute the variance from the formula:
\operatorname{Var}(X)=\sigma_{X}^{2}=E\left[X^{2}\right]\,-\,(E\left[X\right])^{2}E[X]^{2}~=~P\left({ H}\neq1\right)E\left[X^{2}\mid H\neq1\right]+P\left(H=1\right)E\left[X^{2}\mid H=1\right]
= 0.85\mathrm{~*~0}+0.15\int_{0}^{\infty}x^{2}0.1e^{-0.1x}d x
= 0.15\int_{0}^{\infty}0.1x^{2}\ e^{-0.1x}d x
= 0.15(-x^{2} e^{-0.1x}|_{0}^{\infty} \;+\;\textstyle2 {\int_{0}^{\infty}}xe^{-0.1x}d x)
= 0.15\bigl(2*10\int_{0}^{\infty}0.1x\ e^{-0.1x}d x\bigr)
= 2\ *10\ *0.15\Big(\int_{0}^{\infty}0.1x\ e^{-0.1x}d x\Big)
= 20 * 1.5
= 30
\operatorname{Var}(X)=\sigma_{X}^{2}=E\left[X^{2}\right]\,-\,(E\left[X\right])^{2}= 30 – 1.5²
= 27.75
Therefore, as the standard deviation σ_{X} is given by the square root of the variance σ_{X}^{2}= 27.75, we have σ_{X} = √27.75 = 5.27.
The ratio of the standard deviation to the mean (the coefficient of variation) is given by
\sigma_{X}/E[X]=5.27/1.5=3.51