# 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 $p_{\beta,F}(x)=0.$

I will appreciate your effort and answers.

• Welcome to the site. Please be careful when using indentation, and rather avoid them. The software might, as in this case, interpret the subsequent text as "code" and display it as such. – user9072 Nov 12 '14 at 11:13