1
$\begingroup$

To elaborate on the title, here is the entire problem:

Let $X_1, X_2, ..., X_n \thicksim Exp(\lambda)$ be an independent sample.

What's the joint distribution of the sequence of $X_1, X_1 + X_2, ..., X_1 + X_2 + ... + X_{n-1}$ with the condition of $X_1 + X_2 + ... + ... + X_n = t$ ?

And is this joint distribution equal to an $n-1$ member ordered uniform ($Unif[0,t]$) sample's joint distribution, meaning that:

If $Y_1, Y_2, ..., Y_{n-1} \thicksim Unif[0,t]$ independent sample, and we order them: $Y_1^*, Y_2^*, ..., Y_{n-1}^*$, then are these equal:

$$F_{X}(x_1,...,x_{n-1}) = \Bbb{P}(X_1 < x_1, X_1 + X_2 < x_2, ...,~~~ X_1 + X_2 + ... + X_{n-1} < x_{n-1} | X_1 + X_2 + ... + X_n = t) \stackrel{?}{=} \Bbb{P}(Y_1^* < x_1, Y_2 < x_2, ..., Y_{n-1}^* < x_{n-1}) = F_{Y^*}(x_1,...,x_{n-1})$$

where $F_X$ is the joint distribution function of the $X_1, X_2, ...,X_1 + X_2 + ... + X_{n-1}$ sample with the condition of $\sum_{i=1}^n{X_i} = t$ and $F_{Y^*}$ is the joint distribution function of the $Y_1^*, Y_2^*, ..., Y_{n-1}^*$ sample.

If so, prove it; if not, disprove it.

The problem is...:

...that $X_1, X_1 + X_2, ..., X_1 + X_2 + ... + X_n$ aren't independent, so calculating the joint distribution function is hard, especially with a condition.

Ordered samples also follow a Beta distribution, which is generally tough to deal with:

$$\forall k \in \{1,...,n\}: \quad Y_k^* \thicksim \frac{1}{t}Beta(n,n-k+1)$$

Here is what I've tried so far:

1. Introduce new variables: $$A_1 = X_1 \\ A_2 = X_1 + X_2 \\ \vdots \\ A_n = X_1 + X_2 + \dots + X_n$$

This way, we can write up the $X$'s like so: $$X_1 = A_1 \\ X_2 = A_2 - A_1 \\ X_3 = A_3 - A_2 \\ \vdots \\ X_n = A_n - A_{n-1}$$

We can also calculate the individual distributions of these $A$'s:

$$\forall k \in \{1,...,n\} \quad A_k \thicksim Exp\left(\frac{\lambda}{k}\right)$$

But this didn't lead me much further, since we still can't write up the joint distribution functions of $A$'s or $X$'s since they're not independent.

2. I tried thinking outside the box: $X_1 + X_2 + ... + X_k$ could mean the arrival time of a truck, and if they're from an exponential distribution, then their arrival times are expected to be uniform. However, expected value says very little about joint distribution, plus this wouldn't be a very mathematically appropriate proof.

Can anyone lead me on the correct path?

$\endgroup$
  • 1
    $\begingroup$ The joint PDF of $Y_1^*$, $Y_2^*$, $\ldots$, $Y_{n-1}^*$ is $$f_{Y_1^*,...,Y_{n-1}^*}(y_1,..., y_{n-1})=(n-1)!f(y_1)\cdots f(y_{n-1}) = (n-1)! \frac{1}{t^{n-1}} \mathbb 1(0\leq y_1\leq \ldots \leq y_n\leq t).$$ $\endgroup$ – NCh Mar 24 at 11:20
1
$\begingroup$

Let $S_i = X_1 + \ldots + X_i$. $$F_{(S_1, ..., S_n)}(x_1, \ldots, x_n) = \\ \int_{-\infty}^{x_1} f_{X_1}(\tau_1) \int_{-\infty}^{x_2 - \tau_1} f_{X_2}(\tau_2) \cdots \int_{-\infty}^{x_n - \tau_1 - ... - \tau_{n - 1}} f_{X_n}(\tau_n) \, d\tau_n \cdots d\tau_1, \\ \frac {\partial^n} {\partial x_n \cdots \partial x_1} F_{(S_1, ..., S_n)}(x_1, \ldots, x_n) = \\ f_{X_1}(x_1) \frac {\partial^{n - 1}} {\partial x_n \cdots \partial x_2} \int_{-\infty}^{x_2 - x_1} f_{X_2}(\tau_2) \cdots \int_{-\infty}^{x_n - x_1 - \tau_2 - ... - \tau_{n - 1}} f_{X_n}(\tau_n) \, d\tau_n \cdots d\tau_2 = \ldots = \\ f_{X_1}(x_1) f_{X_2}(x_2 - x_1) \cdots f_{X_n}(x_n - x_{n - 1}).$$ For $f_{X_i}(x) = \lambda e^{-\lambda x} [0 < x]$, this gives $$f_{(S_1, ..., S_n)}(x_1, \ldots x_n) = \lambda^n e^{-\lambda x_n} [0 < x_1 < \ldots < x_n].$$ Next, $$f_{S_n}(x) = \mathcal L^{-1} {\left[ \left( \frac \lambda {p + \lambda} \right)^{\!n} \right]} = \lambda^n e^{-\lambda x} \mathcal L^{-1}[p^{-n}] = \frac {\lambda^n} {(n - 1)!} x^{n - 1} e^{-\lambda x} \, [0 < x], \\ f_{(S_1, ..., S_{n - 1}) \mid S_n = t}(x_1, \ldots, x_{n - 1}) = \frac {f_{(S_1, ..., S_n)}(x_1, \ldots, x_{n - 1}, t)} {f_{S_n}(t)} = \\ \frac {(n - 1)!} {t^{n - 1}} \, [0 < x_1 < \ldots < x_{n - 1} < t],$$ which is the same as the pdf of the order statistic $(Y_1^*, \ldots, Y_{n - 1}^*)$.

$\endgroup$
2
+25
$\begingroup$

The discrete version of the problem is easier to solve.

Let $X_1, X_2, \dots, X_n$ be (positive) geometric random variables: if you flip a coin that lands heads with some probability $p>0$, the distribution measures the number of coinflips until the coin lands heads.

Then $X_1, X_1 + X_2, \dots, X_1 + X_2 + \dots + X_n$ measure the number of coinflips until the coin lands heads $1, 2, \dots, n$ times. If we condition on $X_1 + X_2 + \dots + X_n = T$, then we are conditioning on the fact that the $T^{\text{th}}$ coinflip, and exactly $n-1$ of the preceding coinflips, landed heads.

Here, each sequence of $T$ valid coinflips had the same probability $p^n (1-p)^{T-n}$ of occurring before we conditioned on $X_1 + X_2 + \dots + X_n = T$, and so the conditional distribution is the same as the distribution we get by picking, uniformly at random, $n-1$ distinct positions for the first $n-1$ coinflips to land heads.

We can approximate the continuous problem with the discrete one arbitrarily well. Divide the time period $[0,t]$ into $T = \frac{t}{\delta}$ steps, and on each one, flip a coin that lands heads with probability $\lambda\delta$. Let $X_i$ be the interval between the $(i-1)^{\text{th}}$ time and the $i^{\text{th}}$ time the coin lands heads, and condition on the $n^{\text{th}}$ time the coin lands heads being the $T^{\text{th}}$ step.

On one hand, for each $\delta > 0$, we obtain the discrete process described above with $p = \lambda\delta$ and $T = \frac{t}{\delta}$.

On the other hand, as $\delta \to 0$, the $(X_1, X_2, \dots, X_n)$ joint distribution converges in some reasonable sense to the exponential one, and the distribution of $(Y_1, Y_2, \dots, Y_{n-1})$, the distribution of $n-1$ uniformly random time steps, converted to real numbers in $[0,t]$, converges in the same sense to a joint uniform distribution.

(In particular, we can prove convergence of $F_X$ and $F_Y$, which is what we want.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.