Obtaining CDF of a continuous Random Variable

Question

Suppose that $T_i$ and $C_i$ are exponentially distributed of rates $\psi \lambda_i$ and $\lambda_i$ respectively. Letting $Z_i=T_i / C_i$, we obtain: $$ \operatorname{pr}(Z_i>z)=\int_0^\infty \operatorname{pr}(T_i>z c) f_{C_i}(c) \, d c=\lambda_i \int_0^\infty \exp(-\lambda_i \psi z c) \exp(-\lambda_i c) \, d c $$ where $f_{C_i}(c)=\lambda_i e^{-\lambda_i c}$ is the density function of $C_i$ at $c.$ I'm wondering why the expression $\operatorname{pr}(Z_i>z)=\int_0^\infty \operatorname{pr}(T_i>z c) f_{C_i}(c) \, dc$ holds. Intuitively, it seems that we are conditioning on the value of $c$, in that we sum over all possible values of $c$, but is there a rigorous proof-based explanation of this? Thanks.

Answer 1

I will assume independence of $T_i$ and $C_i$. It is not even possible to compute $P(T_i >C_iz)$ without this assumption.

Now $P(T_i >C_iz)=\int 1_{\{(t,c): t>zc\}} dF_{T_i}(t) dC_{i} (c)=\int P(T_i >zc) dF_{C_i}(t)(c)$ by Fubini's Theorem and then fact that the joint distribution of $(T_i,C_i)$ is the product of the distributions of $T_i$ and $C_i$ (by independence).

If $C_i$ is known to have density $f_{C_i}$ then we can write $P(T_i >C_iz)= \int P(T_i >zc) f_{C_i}(t)dc$.