----- > [!proposition] Proposition. ([[Price's model when c=a=1]]) > **a) Write down the special case of the large-$n$ [[master equation]] from [[in-degree distribution of Price's model is a power law]] for $c = a = 1$.** > > This is $p_{q} \begin{cases} > \frac{1}{2} [qp_{q-1} - (q+1)p_{q}] & q > 0 \\ > 1+ \frac{1}{2} p_{0} & q=0. > \end{cases}$ > > **b) Show that the in-degree generating function $g_0(z) = \sum_{q=0}^{\infty} p_q z^q$ for this case satisfies the differential equation $g_0(z) = 1 + \frac{1}{2}(z - 1)[zg'_0(z) + g_0(z)]$.** > > We get $ \begin{align}g_{0}(z)= \sum_{q=0}^{\infty}p_{q}z^{q} = & 1 + \frac{1}{2} \sum_{q=1}^{\infty} p_{q-1} z^{q-1} \cdot z^1 - \frac{1}{2} \sum_{q=0}^{\infty} (q + 1)p_q z^q \\ > &= 1 + \frac{1}{2} z \sum_{q=0}^{\infty} (q + 1)p_q z^{q} - \frac{1}{2} \sum_{q=0}^{\infty} (q + 1)p_q z^q \\ > &= 1 + \frac{1}{2} (z - 1) \left[ \sum_{q=0}^{\infty} q p_q z^{q} + \sum_{q=0}^{\infty} p_q z^q \right] \\ > &= 1 + \frac{1}{2} (z - 1) [zg'_0(z) + g_0(z)]. > \end{align}$ > > **c) Show that the function $h(z) = \frac{z^3 g_0(z)}{(1 - z)^2}$ satisfies $\frac{dh}{dz} = \frac{2z^2}{(1 - z)^3}$** > Compute > $\begin{align} > \frac{dh}{dz} &= \frac{(1 - z)^2(z^3 g'_0(z) + 3z^2 g_0(z)) + 2(1 - z)z^3 g_0(z)}{(1 - z)^4} \\ > &= \frac{z^2}{(1 - z)^3} [(1 - z)(zg'_0(z) + 3g_0(z)) + 2z g_0(z)] \\ > &= \frac{2z^2}{(1 - z)^3}, > \end{align}$ > where we have applied part **b**. > > > **d) Hence find a closed-form solution for the generating function $g_0(z)$. Confirm that your solution has the correct limiting values $g_0(0) = p_0$ and $g_0(1) = 1$.** > > We have $\begin{align*} > \frac{2z^2}{(1 - z)^3} &= \frac{2}{(1 - z)^3} - \frac{4}{(1 - z)^2} + \frac{2}{1 - z} \\ > \int \frac{2z^2}{(1 - z)^3} dz &= \int \left( \frac{2}{(1 - z)^3} - \frac{4}{(1 - z)^2} + \frac{2}{1 - z} \right) dz \\ > h(z) &= \frac{1}{(1 - z)^2} - \frac{4}{1 - z} - 2\ln(1 - z) + \text{C}. > \end{align*} > $ > > > **e) Thus find a value for the mean in-degree of a node in Price's model. Is this what you expected?** > > We compute $g_{0}'(1)$: > $\begin{align*} > \frac{dg_0}{dz} &= \frac{1}{2} \left[3z^2(8z - 4 - 4(1 - z)\ln(1 - z)) \right. \\ > &\quad \left. - 3z^2 \left(3z^2 - 2z + 2(1 - z)^2\ln(1 - z)\right)\right]. > \end{align*}$ > and letting $z \to 1$ this gives $c=1$ as expected. > [!proof]- Proof. ([[expected in-degree of a node in Price's model]]) > ~ ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```