diff --git a/_images/kbarfig.png b/_images/kbarfig.png new file mode 100644 index 0000000..f9f8626 Binary files /dev/null and b/_images/kbarfig.png differ diff --git a/_sources/notebooks/kbar.ipynb b/_sources/notebooks/kbar.ipynb index 4e00625..cb6e48d 100644 --- a/_sources/notebooks/kbar.ipynb +++ b/_sources/notebooks/kbar.ipynb @@ -79,7 +79,7 @@ "id": "8a30128d", "metadata": {}, "source": [ - "Re-arranging to solve for the $\\kappa=\\bar kappa$ at which this holds as an equality, we find:\n", + "Re-arranging to solve for the $\\kappa=\\bar \\kappa$ at which this holds as an equality, we find:\n", "\n", "Setting $c_{1}^{0} =c_1^F = c_{2}^{0}$ equal to the implied efficient continuation contract into no-renegotiation constraint we can solve for the $\\bar \\kappa$ value of $\\kappa$ that allows this to just hold:\n", "\n", @@ -104,6 +104,17 @@ "For the monopoly case we simply replace the competitive $c_1^F$ term by the analogous $c_1^{mF}$ efficient monopoly contract ters. The monopoly formula for the threshold is thus $\\bar \\kappa^m = c_1^{mF} \\cdot \\Upsilon$" ] }, + { + "cell_type": "markdown", + "id": "241fe6e2", + "metadata": {}, + "source": [ + "This diagram helps to visualize. Zero's first-best commitment smoothing contract at $F$ is not because the $\\kappa =\\bar \\kappa$ renegotiation cost is just high enough to make the bank reject the most favorable renegotiated contract that could be offered to the bank at $R$ .\n", + "\n", + "\n", + "" + ] + }, { "cell_type": "markdown", "id": "20ae35ed-0ed9-4ad6-a94c-cde7fab05407", @@ -261,14 +272,14 @@ }, { "cell_type": "code", - "execution_count": 7, + "execution_count": 8, "id": "7279cf5b-f088-4152-b8d2-2d31d686c478", "metadata": {}, "outputs": [ { "data": { "application/vnd.jupyter.widget-view+json": { - "model_id": "38c24a8cd8324e2697fabb295b069ea0", + "model_id": "db9e21b916cd48458439fbb7c7ef2a23", "version_major": 2, "version_minor": 0 }, @@ -281,8 +292,16 @@ } ], "source": [ - "interact(plotkb, rho=(0.1, 2, 0.101), y0=(50,200,10));" + "interact(plotkb, rho=(0.1, 2, 0.101), y0=(50,300,10));" ] + }, + { + "cell_type": "code", + "execution_count": null, + "id": "e87f6278", + "metadata": {}, + "outputs": [], + "source": [] } ], "metadata": { diff --git a/notebooks/Contract_plots.html b/notebooks/Contract_plots.html index 044468b..a4f8a1a 100644 --- a/notebooks/Contract_plots.html +++ b/notebooks/Contract_plots.html @@ -63,7 +63,7 @@ - + @@ -481,7 +481,7 @@
The parameters (\(\beta = 0.4\), \(\rho = 0.8\)) used were chosen to exagerate curvature and spacing for presentation clarity but the essential relationships hold for more reasonable assumptions. In the interactive further below you can vary the parameters yourself using sliders.
A gif to illustrate how the plot adjusts to parameter changes:
diff --git a/notebooks/kbar.html b/notebooks/kbar.html index 82f7937..5de47c8 100644 --- a/notebooks/kbar.html +++ b/notebooks/kbar.html @@ -63,7 +63,7 @@ - + @@ -439,7 +439,7 @@Re-arranging to solve for the \(\kappa=\bar kappa\) at which this holds as an equality, we find:
+Re-arranging to solve for the \(\kappa=\bar \kappa\) at which this holds as an equality, we find:
Setting \(c_{1}^{0} =c_1^F = c_{2}^{0}\) equal to the implied efficient continuation contract into no-renegotiation constraint we can solve for the \(\bar \kappa\) value of \(\kappa\) that allows this to just hold:
The above is for the competitive case.
For the monopoly case we simply replace the competitive \(c_1^F\) term by the analogous \(c_1^{mF}\) efficient monopoly contract ters. The monopoly formula for the threshold is thus \(\bar \kappa^m = c_1^{mF} \cdot \Upsilon\)
+This diagram helps to visualize. Zero’s first-best commitment smoothing contract at \(F\) is not because the \(\kappa =\bar \kappa\) renegotiation cost is just high enough to make the bank reject the most favorable renegotiated contract that could be offered to the bank at \(R\) .
+Several functions used can be found in the python module Contract.py
@@ -540,12 +542,12 @@interact(plotkb, rho=(0.1, 2, 0.101), y0=(50,200,10));
+interact(plotkb, rho=(0.1, 2, 0.101), y0=(50,300,10));