What is the Variance of $(1-X)/X$ and the Expectation of $(1-X)/X^{2}$?

Solution 1:

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.