# Finding a general form of the density function when we have a four dimensional random variable

Consider a subject having time of the specific event $T_i$, which is a single sample from a distribution $F_i$ with density $f_i$ and support $[t_{\min},t_{\max}]$, for $i= 1,\ldots,n$. Let these subjects be interviewed at times $S_1,\ldots,S_n \in [t_{\min},t_{\max}]$, respectively. Let $\varepsilon_i$ be the indicator of recalling the age at that event and $\delta_i=I_{(T_i\leqslant S_i)}$. Let $V_i=(S_i-T_i)\varepsilon_i\delta_i,$ and $Y_i=(S_i,V_i,\delta_i)$. Its clear that $Y_i$ is non negative random variable which had density with respect to proper measure. Let $t_1<\ldots<t_{j}<\ldots<t_{n_2}$ are observed points of random variable $T_i$ $(n_2<n)$ and $$\bar{F}_i(t)=[\bar{F}(t)]^{\exp(\beta^TZ_i)},$$ where $\bar F=1-F$.

The joint density of $X_i=(Y_i,Z_i)$ is $$p_{\beta,F}(x_i)=\sum_{j=1}^{n_2} \alpha_{ij}\big[\bar F(t_{j})^{e^{z_i\beta}}-\bar F(t_{j+1})^{e^{z_i\beta}}\big]g(z_i)$$ where $g$ is the density of $Z$ and $$\alpha_{ij} = \left\{ \begin{array}{ll} I\big(t_j\in (s_i,t_{max}]) & \mbox{if}~~ \delta_i=0,\\ \mbox{Constant}.I\big(t_j\in [T_i,T_i]) & \mbox{if}~~ \delta_i=1,\ (s_i-T_i)>0,\\ \mbox{Constant} .I\big(t_j\in ((s_i-x_l,s_i-x_{(l-1)}]) & \mbox{if}~~\delta_i=1,\ (s_i-T_i)=0.\\ \end{array} \right.$$ if $x_{l-1}< x_l,\ l=1,2,...,7$ are fixed ordered points.

The question is what is the general form of the density of $X$? In other word what is $p_{\beta,F}(y,z)$?

I tried to write the probability in the following form $$p_{\beta,F}(x)=\int \alpha(y,t')d\left(1-\bar F(t')^{\exp(\beta^Tz)}\right)\times g(z).$$ where $$\alpha(y,t') = \left\{ \begin{array}{ll} I\big(t'\in (s,t_{max}]) & \mbox{if}~~ \delta=0,\\ \mbox{Constant}.I\big(t'=s+u) & \mbox{if}~~ \delta=1,\ v>0,\\ \mbox{Constant} .I\big(t'\in ((s-x_l,s-x_{(l-1)}]) & \mbox{if}~~\delta=1,\ v=0.\\ \end{array} \right.$$ if $x_{l-1}< x_l,\ l=1,2,...,7$ are fixed ordered points. But the problem is if the distribution of $T$ is continuous by the second form of $\alpha(y,t')$, we will have $p_{\beta,F}(x)=0.$