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Question

X-Bacterias in a petri dish abide to the following rules:

  1. each bacteria evolves identically and independently from the others.
  2. each bacteria is replaced by four new bacterias after a random time with $e(\beta)$-distribution, where $\beta = 1$ hours$^{−1}$.

Denote $X_t$ by the number of bacterias at time t and assume that $X_0 = n$, where $n \ge 1$.

a) Show that $\mathbb E[X_t \mathbb{1}_{T_1\le t}]=e^{-\beta t}\int_{0}^{t}4\beta e^{\beta s}\mathbb E[X_s]ds$

b) Deduce that $\mathbb E[X_t]=e^{-\beta t}\int_{0}^{t}4\beta e^{\beta s}\mathbb E[X_s]ds +e^{-\beta t}$

c) Show that $\mathbb E[X_t]=e^{3\beta t}$

My attempt's

a) I am unsure how to deal with the indicator. First instinct is to say $\mathbb E[X_t\mathbb{1}_{T_1\le t}]=\mathbb E[X_t |T_1\le t]$, which I believe is incorrect, even still, then using conditional expectation on this is tricky.

b) Feel like this requires part a) so haven't attempted this as of yet.

c) As for this, assuming I have proved b),

Let $\mu(t):=E[X_t]=e^{-\beta t}\int_{0}^{t}4\beta e^{\beta s}\mathbb E[X_s]ds +e^{-\beta t}$

Now computing the following, using Leibniz's integral rule: $\mu'(t)=\frac{d}{dt}\{\int_{0}^{t}4\beta e^{\beta (s-t)}\mu(s)ds+e^{-\beta t}\}$, we get...

$\mu'(t)=\frac{d}{dt}(t)\cdot4[\beta e^{\beta (s-t)}\mu(s)]_{s=t}-0+\int_{0}^{t}\partial_t\{4\beta e^{\beta (s-t)}\mu(s)\}ds-\beta e^{\beta s}=4\beta\mu(t)-\beta\int_{0}^{t}4\beta e^{\beta (s-t)}\mu(s)ds-\beta e^{\beta s}=4\beta\mu(t)-\beta[\int_{0}^{t}4\beta e^{\beta (s-t)}\mu(s)ds+e^{-\beta t}]=4\beta\mu(t)-\beta\mu(t)=3\beta\mu(t)$

Hence, we have the following differential equation:$\mu'(t)=3\beta\mu(t)$

$\therefore \mu(t)=\mathbb E[X_t]=e^{3\beta t}$

Comments

Any hints with a) would be greatly appreciated, I feel like I am missing some crucial steps and haven't made much progress over the past few hours. Moreover, some alternative solutions would be interesting to see also:)

George Cooper
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2 Answers2

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We start with $n$ bacteria, each following its own reproduction path. So we concentrate on a lone bacterium. We have that $X_0=1$ and $X_{T_1}=4$. At that point, each of the $4$ bacteria follows its own path, and so on. Clearly $$E[X_t\mathbf{1}_{\{t<T_1\}}]=E[X_t|t<T_1]P(t<T_1)=X_0e^{-\beta t}=e^{-\beta t}$$ Notice that on $t\geq T_1$ we have $X_t= X^{(1)}_{t-T_1}+...+X^{(4)}_{t-T_1}$. This is because we have four bacteria after the event $T_1$ which 'restart' at $T_1$. Then $$\begin{aligned}E[X_t\mathbf{1}_{\{t\geq T_1\}}]&=4E[E[X_{t-T_1}^{(1)}|T_1]\mathbf{1}_{\{t\geq T_1\}}]=\\ &=4\int_{[0,t]}E[X_{t-s}]\beta e^{-\beta s}ds=\\ &=4\int_{[0,t]}E[X_{u}]\beta e^{-\beta (t-u)}du\end{aligned}$$ So clearly $$E[X_t]=E[X_t\mathbf{1}_{\{t\geq T_1\}}]+E[X_t\mathbf{1}_{\{t<T_1\}}]=4\int_{[0,t]}E[X_{u}]\beta e^{-\beta (t-u)}du+e^{-\beta t}$$ Quickly, using the ansatz $e^{3\beta t}$ (innocently provided by the answer): $$4\beta e^{-\beta t}\int_{[0,t]} e^{4\beta u}du+e^{-\beta t}=e^{-\beta t}(e^{4\beta t}-1)+e^{-\beta t}$$ which shows $E[X_t]= e^{3\beta t}$.

Snoop
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  • Wow, beautifully done. Is there anyway you could explain to me $E[X_t\mathbf{1}_{{t<T_1}}]=E[X_t|t<T_1]P(t<T_1)$ Other than that you have made this as clear as day to me. Thank you :) @Snoop – George Cooper Feb 17 '22 at 22:48
  • You're welcome @GeorgeCooper That equality is a property of conditional expectation! I do recall there are proofs here on math SE. Look for it on Approach Zero – Snoop Feb 17 '22 at 22:52
  • Oh I see, yeah $E[X|A]=\frac{E[X\mathbb 1_{A}]}{\mathbb P(A)}$ by definition. Great! Thanks again – George Cooper Feb 17 '22 at 22:58
  • I have only have seen this in a discrete context. Hadn't thought to use it in a continuous one! – George Cooper Feb 17 '22 at 23:00
  • @GeorgeCooper had to fix a passage – Snoop Feb 17 '22 at 23:29
  • I am confused with this step: $E[X_t\mathbf{1}{{t\geq T_1}}]=4E[E[X{t-T_1}|T_1]\mathbf{1}_{{t\geq T_1}}]$ – George Cooper Feb 17 '22 at 23:36
  • I am unsure how to justify this? – George Cooper Feb 17 '22 at 23:36
  • @GeorgeCooper hope it is clear now with more notation. $E[X^{(1)}_{t-T_1}|T_1=s]=E[X_{t-s}]$ because the new bacteria are identical copies of the initial bacterium, but their evolution is not influenced by $T_1$, which influences only the time at which we take the expectation. – Snoop Feb 18 '22 at 00:15
  • Thank you that is a lot clearer now. That makes a lot more sense :) – George Cooper Feb 18 '22 at 15:01
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Here is another approach you may find useful. It's a bit backwards, but illustrates a nice strategy nevertheless.

Fix $t\geq 0$ and let $N_t$ represent the number of times the population of bacteria quadrupled in $[0,t]$. Then $N_t\sim \text{Poisson}(\beta t)$ and $X_t=n4^{N_t}$. Now use LOTUS: $$\begin{eqnarray*}\mathbb{E}(X_t)&=&\sum_{k=0}^{\infty}n4^{k}\cdot \mathbb{P}(N_t=k) \\ &=&\sum_{k=0}^{\infty}n4^k\cdot e^{-\beta t}\frac{(\beta t)^k}{k!} \\ &=& ne^{-\beta t}\sum_{k=0}^{\infty}\frac{(4\beta t)^k}{k!} \\ &=& n e^{-\beta t}e^{4\beta t} \\ &=& ne^{3\beta t}\end{eqnarray*}$$ This shows item (c). From total law of expectation we have $$\begin{eqnarray*}\mathbb{E}(X_t) &=&\mathbb{E}(X_t\cdot 1_{\{T_1 \leq t\}})+\mathbb{E}(X_t|T_1>t)\mathbb{P}(T_1>t) \end{eqnarray*}$$ Because $\mathbb{E}(X_t)=ne^{3\beta t}, \mathbb{E}(X_t|T_1>t)=n$, and $\mathbb{P}(T_1>t)=e^{-\beta t}$ we have that $$ne^{3\beta t}=\mathbb{E}(X_t\cdot 1_{\{T_1 \leq t\}})+ne^{-\beta t}$$ This is equivalent to saying $$\mathbb{E}(X_t\cdot 1_{\{T_1 \leq t\}})=n\Big(e^{3 \beta t}-e^{-\beta t}\Big)=e^{-\beta t}\int_0^t4\beta e^{\beta s}\mathbb{E}(X_s)\mathrm{d}s$$ This justifies (b). Using the formula we established in (c) you can also justify (b) easily.