Find $c$ such that the test which rejects when $X>c$

Question

Consider a random variable $Z$ having pdf $f(z)=\frac{1}{2} e^{-|z-\mu|}$ , where $z$ is real.We observe $X=max(0,Z)$. Find $c$ such that the test which rejects when $X>c$ has size $0.05$ under $H_0:\mu=0$.

Now, we know: $P_{H_0}(X>c)=P_{H_0}(0>c)P(Z<0) +P_{H_0}(Z>c)P(Z \ge 0)=0.05$ Since $c$ is positive otherwise the test will have size $1$, so $P_{H_0}(c<0)=0$.Thus, we have $ P_{H_0}(Z>c)\frac{1}{2}=0.05$ which gives $c=\ln 5$. But I am getting to know that the value of $c=\ln 10$. So, where did I go wrong?

Answer 1

If you use the partition $\{Z\geq 0,\,Z<0\}$, $$P(X>c) = P(X>c,Z\geq0) + P(X>x,Z<0) = P(X>c,Z\geq 0)= P(Z>c)$$

Hence, you have that $X>c$ iff $Z>c$ (if $c>0$), so $$P(X>c) = \int_c^\infty \frac{e^{-z}}{2}\mathrm d z = \frac{e^{-c}}{2}.$$