Compute Quantile Function that uses an Indicator function

Question

I have to calculate the Quantile Function, $$ F^{-1}_{L}(1-0.05), $$ where the distribution is given by $L=-(1-\textbf{1})$. Here $\textbf{1}$ is the indicator function with $P(\textbf{1}=1)=0.02$. Thus, inserting this we get, $$ F^{-1}_{L}(1-0.05)=F^{-1}_{-(1-\textbf{1})}(0.95) = F^{-1}_{(\textbf{1}-1)}(0.95). $$ Using the fact that $F^{-1}_{g(X)}(p)=g(F^{-1}_{X}(p))$ should yield that $F^{-1}_{(\textbf{1}-1)}(0.95)=F^{-1}_{\textbf{1}}(0.95)-1.$ Because the distribution function $F_{\textbf{1}}(1)=P(\textbf{1}\leq1)=1$, due to the indicator function only taking values $1$ or $0$, we get $F^{-1}_{\textbf{1}}(0.95)=1$. Thus, $$ F^{-1}_{L}(1-0.05)=F^{-1}_{\textbf{1}}(0.95)-1=1-1=0. $$ However, the short solution that I have says that $F^{-1}_{L}(1-0.05)=-1$. Could someone please show me how they arrived to that? What am I doing wrong?

Answer 1

Your steps were correct all the way up until $F_{\mathbf{1}}^{-1}(0.95)=1$. This is actually $0$, since $P(\mathbf{1} \leq 0) = 0.98 \geq 0.95$, while $P(\mathbf{1} \leq a)=0$ for all $a<0$. Making this change gives the expected result.