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Let $X_t$ be a simple birth process (with birth rate $\lambda_n = n\lambda$) such that $X_0 = 1$. Set $\mu_t = E[X_t]$ and $\nu_t = E[X_t^2]$. Find differential equations for $\mu_t$ and $\nu_t$ and hence calculate the variance of $X_t$.

I've tried using the Kolmogorov forward differential equation to get $\mu_t$ which gave me $e^{-\lambda t}$ but I'm not sure how to calculate $\nu_t$.

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  • $\begingroup$ "with birth rate λn = λn" - come again? $\endgroup$
    – Math1000
    May 25 '18 at 17:22
  • $\begingroup$ that should be just λn, sorry. $\endgroup$
    – J.P.D.W
    May 25 '18 at 17:25
  • $\begingroup$ So the birth rate is constant? $\endgroup$
    – Math1000
    May 25 '18 at 18:00
  • $\begingroup$ no, its increasing as the population increases $\endgroup$
    – J.P.D.W
    May 25 '18 at 18:08
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The differential equations (with initial conditions) for this Yule process with linear birth rates are \begin{align} P_1'(t) &= -\lambda P_1(t),\quad P_1(0)=1\tag 1\\ P_n'(t) &= -n\lambda P_n(t)+(n-1)\lambda P_{n-1}(t),\quad P_n(0)=0.\\ \end{align} Solving $(1)$ yields $P_1(t) = e^{-\lambda t}$. Assume now that $P_n(t) = e^{-\lambda t}(1-e^{-\lambda t})^{n-1}$. Then \begin{align} P_{n+1}'(t) &= -(n+1)\lambda P_{n+1}(t) + n\lambda P_n(t)\\ &= -(n+1)\lambda P_{n+1}(t) + n\lambda e^{-\lambda t}(1-e^{-\lambda t})^{n-1}, \end{align} so $$ P_{n+1}'(t) +(n+1)\lambda P_{n+1}(t) = n\lambda e^{-\lambda t}(1-e^{-\lambda t})^{n-1}.\tag2 $$ This is a first-order linear differential equation, with integrating factor $$ \mu(t) = e^{\int (n+1)\lambda \ \mathsf dt} = e^{(n+1)\lambda t}. $$ Multiplying $(2)$ by $\mu(t)$, we have $$ \frac{\mathsf d}{\mathsf dt}[P_{n+1}(t)e^{(n+1)\lambda t}] = n\lambda e^{n\lambda t}(1-e^{-\lambda t})^{n-1}. $$ Integrating and multiplying by $1/\mu(t)$ yields $$ P_{n+1}(t) = e^{-\lambda t}(1-e^{-\lambda t})^n,\tag3 $$ so by induction $(3)$ holds for all $n\geqslant 1$. The first moment of $X(t)$ is given by \begin{align} \mathbb E[X(t)] &= \sum_{n=1}^\infty n P_n(t)\\ &= \sum_{n=1}^\infty ne^{-\lambda t}(1-e^{-\lambda t})^{n-1}\\ &= e^{\lambda t}, \end{align} and the second moment by \begin{align} \mathbb E[X(t)^2] &= \sum_{n=1}^\infty n^2 P_n(t)\\ &= \sum_{n=1}^\infty n^2e^{-\lambda t}(1-e^{-\lambda t})^{n-1}\\ &= e^{\lambda t} \left(2 e^{\lambda t}-1\right), \end{align} so the variance is \begin{align} \mathsf{Var}(X(t) &= E[X(t)^2] - E[X(t)]^2\\ &= e^{\lambda t} \left(2 e^{\lambda t}-1\right) - (e^{\lambda t})^2\\ &= e^{\lambda t} \left(e^{\lambda t}-1\right). \end{align}

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  • $\begingroup$ how do you go from the second line of E[X(t)] to the 3rd line? $\endgroup$
    – J.P.D.W
    May 25 '18 at 20:40
  • $\begingroup$ \begin{align} \sum_{n=1}^\infty nz^{n-1} &= \sum_{n=0}^\infty (n+1)z^n\\ &= \frac{\mathsf d}{\mathsf dz} \sum_{n=0}^\infty z^n\\ &= \frac{\mathsf d}{\mathsf dz} \left[ \frac1{1-z}\right]\\ &= \frac1{(1-z)^2}. \end{align} $\endgroup$
    – Math1000
    May 25 '18 at 21:14
  • $\begingroup$ Looks like I had a couple errors in the computations, fixed now. $\endgroup$
    – Math1000
    May 25 '18 at 21:20

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