# How can I find the distribution function of the following random variables?

Suppose that the random variables $$Z_i$$ are defined as follows:

$$$$Z_i = D(0, t_i)(R_{i-1} +c)\Delta N,$$$$ where $$D(0, t_i)= \exp\{-\int_{0}^{t_i} r_u du\}$$ for which $$r_u$$ follows a CIR model, $$R_{i-1}$$ stands for the forward LIBOR rate, and $$c$$, $$\Delta$$, $$N$$ are positive constants. I am looking for the distribution function of $$Z_i$$. However, we already know the fact that $$r_u$$ has a non-central Chisquare distribution. My main issue is how to get rid of the integral on the left side of the inequality below. I think there is something wrong. I am not sure whether we need to take into account a specific rule when we want to take differential of both sides of an inequality. Let me share my solution. For $$y> 0$$, we have that

$$$$\mathbb{P}(Z_i \leq y) = \mathbb{P}\Big(D(0, t_i)(R_{i-1} +c)\Delta N \leq y\Big) = \mathbb{P}\Big(D(0, t_i) \leq \frac{y}{(R_{i-1}+c)\Delta N}\Big) = \mathbb{P}\Big(\exp\{-\int_{0}^{t_i} r_u du\} \leq \frac{y}{(R_{i-1}+c)\Delta N}\Big) = \mathbb{P}\Big(\int_{0}^{t_i} r_u du \geq \log\Big(\frac{(R_{i-1}+c)\Delta N}{y}\Big)\Big) = \mathbb{P}\Big(d\left[\int_{0}^{t_i} r_u du\right]\geq d\left[\log\Big(\frac{(R_{i-1}+c)\Delta N}{y}\Big)\right]\Big) = \mathbb{P}\Big(r_{t_i}dt_i\geq d\left[\log\Big(\frac{(R_{i-1}+c)\Delta N}{y}\Big)\right]\Big) = \mathbb{P}\Big(r_{t_i}\geq \frac{d}{dt_i}\left[\log\Big(\frac{(R_{i-1}+c)\Delta N}{y}\Big)\right]\Big) = \mathbb{P}\Big(r_{t_i}\geq \frac{R^{\prime}_{i-1}}{R_{i-1}+c}\Big)$$$$ where the last line is obtained by knowing $${[\log(f)]}^{\prime} = \frac{f^{\prime}}{f}$$. Note here that the LIBOR rate is defined as follows: $$$$R_{i-1} = \frac{1}{t_i - t_{i-1}}\Big(\frac{P(t_{i-1}, t_{i-1})}{P(t_{i-1}, t_i)} -1\Big) = \frac{1}{\Delta}\Big(\frac{1}{P(t_{i-1}, t_{i})} -1\Big)$$$$ where $$P(t_{i-1}, t_i)$$ denotes the zero-coupon bond price issuing at time $$t_{i-1}$$ and maturing at time $$t_i$$ with following solution

$$$$P(t_{i-1}, t_i) = A(t_{i-1}, t_i)\exp\Big\{-B(t_{i-1}, t_i)r(t_{i-1})\Big\}$$$$

, $$t_{i} - t_{i-1} = \Delta$$, and finally the notation $$d[.]$$ represents the derivative of a given term. Moreover, $$A(., .)$$ and $$B(., .)$$ are deterministic functions of times $$t_{i-1}$$ and $$t_i$$.

As you can see, there is no $$y$$ left at the end. I doubt that the final result is showing a distribution function because we have a constant, not a function of $$y$$.