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

\begin{equation} Z_i = D(0, t_i)(R_{i-1} +c)\Delta N, \end{equation} 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

\begin{equation} \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) \end{equation} 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: \begin{equation} 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) \end{equation} 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

\begin{equation} 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\} \end{equation}

, $t_{i} - t_{i-1} = \Delta$, 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 $y$.

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