Let

E_{n}

be the holding time in state

n

. The variables are independent with

E_{n} \sim Exp (nλ)

, and the explosion time would be

T_{\infty} = \sum_{n \geq 1} E_{n} .

For

s > 0

E e^{- s T_{\infty}} = \prod_{n \geq 1} \frac{nλ}{nλ + s} .

This product is zero because

\sum_{n} lo g (1 + s / (nλ)) = \infty

. If

T_{\infty}

were finite with positive probability, the nonnegative variable

e^{- s T_{\infty}}

would have positive expectation. Therefore

T_{\infty} = \infty

almost surely, and the Yule process is nonexplosive.

ii

Solution

For ordered birth times

0 < s_{1} < \dots < s_{n} < t

, with

s_{0} = 0

, the joint density of births at those times followed by no further birth before

t

[\prod_{k = 1}^{n} kλ e^{- kλ (s_{k} - s_{k - 1})}] e^{- (n + 1) λ (t - s_{n})} = n! λ^{n} e^{λ \sum_{k} s_{k} - (n + 1) λ t} .

Also

P (X_{t} = n + 1) = e^{- λ t} (1 - e^{- λ t})^{n} = e^{- (n + 1) λ t} (e^{λ t} - 1)^{n} .

n! \prod_{k = 1}^{n} \frac{λ e ^{λ s_{k}}}{e ^{λ t} - 1},

which is exactly the joint density of the order statistics of

n

independent variables with the stated density. This proves the conditional Yule birth times result.

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