Birth processes, X-Bacterias in a petri dish abide to the following rules...

Question

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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 $\mathbb E[X_t]=e^{3\beta t}$

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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:$\mu'(t)=3\beta\mu(t)$

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

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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:)

Answer 1

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 $E[X_t]= e^{3\beta t}$.

Answer 2

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.