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Knowing that $X = \begin{cases} 0.05 \ \ \text{with probability} \ \ 0.7\\ 0.2\ \ \text{with probability} \ \ 0.3\end{cases}$

What would be the Variance of $\frac{1-X}{X}$ and the Expectation of $\frac{1-X}{X^2}$ ?

I have computed it and would like to know if my development is correct : \begin{aligned} Var\left[\frac{1-X}{X}\right] &= E\left[\left(\frac{1-X}{X}\right)^2\right] - \left(E\left[\frac{1-X}{X}\right]\right)^2 \\ &= \left(\frac{0.95}{0.05}\right)^2\cdot 0.7 + \left(\frac{0.8}{0.2}\right)^2\cdot 0.3 - \left(\frac{0.95}{0.05}\cdot 0.7 + \frac{0.8}{0.2}\cdot 0.3\right)^2 = 47.25 \end{aligned}

and

$$E\left[\left(\frac{1-X}{X^2}\right)\right] = \left(\frac{0.95}{0.05^2}\right)\cdot 0.7 + \left(\frac{0.8}{0.2^2}\right)\cdot 0.3 = 272 $$

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    $\begingroup$ Your method is alright. I have not checked the computation. $\endgroup$
    – joy
    Dec 21, 2021 at 19:07
  • $\begingroup$ The variance and expectation you found are indeed correct. $\endgroup$
    – user170231
    Dec 21, 2021 at 19:09

1 Answer 1

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Since $$\operatorname{E}\left[\frac{1-X}{X}\right]=\sum_{i=1}^2 \frac{1-x_i}{x_i}p_i$$ $$=\frac{0.95}{0.05}\cdot 0.70+\frac{0.80}{0.20}\cdot 0.30=14.5$$ And $$\operatorname{E}\left[\left(\frac{1-X}{X}\right)^2\right]=\sum_{i=1}^2 \left(\frac{1-x_i}{x_i}\right)^2 p_i$$ $$=\left(\frac{0.95}{0.05}\right)^2\cdot 0.70+\left(\frac{0.80}{0.20}\right)^2\cdot 0.30$$ $$=361\cdot 0.70+16\cdot 0.30=257.5$$ We have $$\operatorname{Var}\left(\frac{1-X}{X}\right)=257.5-14.5^2=47.25$$ Also $$\operatorname{E}\left[\frac{1-X}{X^2}\right]=\sum_{i=1}^2 \frac{1-x_i}{(x_i)^2}p_i$$ $$=\frac{0.95}{0.05^2}\cdot 0.70+\frac{0.80}{0.20^2}\cdot 0.30=272$$ So yes, as other users already mentioned, your methods and answers are indeed correct.

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