Let's assume we only have right-censoring after a fixed time $\tau$, then for i.i.d. exponentially distributed failure times $X_i \sim \text{Exp}(\lambda)$ the sufficient statistics (under the clearly fulfilled assumption of random right-censoring) are $y=\sum_{i=1}^N 1\{ x_i \leq \tau \}$ the number of subjects with an event prior to censoring and $t=\sum_{i=1}^N \min(x_i,\tau)$ the total observed follow-up to first event or censoring. Are there some known results or closed form distribution function for the joint distribution for $Y$ and $T$ similar to how we know that the sum of exponentials (without censoring) is Erlang distributed or how in this particular case $Y \sim \text{Bin}(n, 1-e^{-\lambda \tau})$?

In fact, when I try to get the cumulative distribution function for $T$ I already get extremely messy stuff, but perhaps I just overlook some clever ways to simplify. It would be pretty useful to figure this out, since it then allows simpler simulation of data, as well as mathematical evaluation of the properties of certain evaluation methods.


1 Answer 1


As in your question, let $X_1,...,X_n \sim \text{IID Exp}(\lambda)$ and for some $\tau > 0$ define the statistics:

$$Y \equiv \sum_{i=1}^n \mathbb{I}(X_i \leqslant \tau) \quad \quad \quad T \equiv \sum_{i=1}^n \min(X_i , \tau).$$

You want to find the joint distribution of $(Y,T)$. Firstly, we note that it is trivial to establish the marginal distribution $Y \sim \text{Bin}(n, 1-e^{-\lambda \tau})$, so finding the joint distribution requires you to find the conditional distribution of $T$ given $Y$. Conditional on $Y=y$ we can write $T$ as:

$$T = (n-y) \tau + S \quad \quad \quad S \equiv \sum_{i=1}^{y} X_{(i)}$$

and the latter is equivalent to $S \overset{\text{Dist}}{\sim} \sum_{i=1}^{y} X_i^*$ where $X_1^*,...,X_y^* \sim \text{IID TrExp}(\lambda, \tau)$ are independent truncated exponential random variables. Thus, the only difficult part of the question is to find the distribution of the sum of truncated exponential distributions.

Finding the distribution of a sum of truncated exponential random variables: For an arbitrary $X^* \sim \text{TrExp}(\lambda, \tau)$ we have the density:

$$p_{X^*}(x) = \frac{e^{-\lambda x}}{1-e^{-\lambda \tau}} \cdot \mathbb{I}(0 \leqslant x \leqslant \tau).$$

This distribution has characteristic function:

$$\varphi_{X^*}(t) \equiv \mathbb{E}(\exp(itX^*)) = \int \limits_0^{\tau} \frac{e^{-(\lambda - it) x}}{1-e^{-\lambda \tau}} dx = \frac{1}{\lambda-it} \cdot \frac{1-e^{-(\lambda-it) \tau}}{1-e^{-\lambda \tau}}.$$

Hence, the characteristic function for $S$ is:

$$\begin{equation} \begin{aligned} \varphi_S(t) = \varphi_*(t)^y &= \frac{1}{(\lambda-t)^y} \cdot \frac{(1-e^{-(\lambda-t) \tau})^y}{(1-e^{-\lambda \tau})^y}. \end{aligned} \end{equation}$$

Applying an inverse Fourier transform we obtain the density:

$$\begin{equation} \begin{aligned} p_S(s) &= \frac{1}{2 \pi} \cdot \frac{1}{(1-e^{-\lambda \tau})^y} \int \limits_{-\infty}^{\infty} e^{-its} \cdot \Big( \frac{1-e^{-(\lambda-t) \tau}}{\lambda-t}\Big)^y dt. \end{aligned} \end{equation}$$

I do not see any obvious closed form for this density, so I will leave it in this form.

From here we have the overall joint distribution:

$$\begin{equation} \begin{aligned} p(Y=y,T=t) &= p_Y(y) \cdot p_S(t-(n-y) \tau) \\[6pt] &= {n \choose y} e^{- (n-y) \lambda \tau} \frac{1}{2 \pi} \int \limits_{-\infty}^{\infty} e^{-it'(t-(n-y) \tau)} \cdot \Big( \frac{1-e^{-(\lambda-t') \tau}}{\lambda-t'}\Big)^y dt'. \\[6pt] \end{aligned} \end{equation}$$


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