diff --git a/Exercise sheet 8/exercise_sheet_08.ipynb b/Exercise sheet 8/exercise_sheet_08.ipynb
index 2740717..b69cf70 100644
--- a/Exercise sheet 8/exercise_sheet_08.ipynb	
+++ b/Exercise sheet 8/exercise_sheet_08.ipynb	
@@ -421,7 +421,38 @@
     }
    },
    "source": [
-    "YOUR ANSWER HERE"
+    "**To proof**\n",
+    "\n",
+    "$\\frac{1}{V}\\mathbb{E}[ \\rho_T(r) ] = \\mathbb{P}(d_T(X,Y) = r)$\n",
+    "\n",
+    "**Proof**\n",
+    "\n",
+    "$$\n",
+    "\\frac{1}{V} \\mathbb{E}\\left[ \\rho_T(r)\\right]\n",
+    "    = \\frac{1}{V} \\mathbb{E} \\left[\\frac{1}{V} \\sum_{x=0}^{V-1}\\sum_{y=0}^{V-1} \\mathbf{1}_{\\{d_T(x,y)=r\\}} \\right]\n",
+    "    = \\frac{1}{V^2} \\mathbb{E} \\left[ \\sum_{x=0}^{V-1}\\sum_{y=0}^{V-1} \\mathbf{1}_{\\{d_T(x,y)=r\\}} \\right]\n",
+    "    = \\frac{1}{V^2} \\sum_{x=0}^{V-1}\\sum_{y=0}^{V-1} \\mathbb{E} \\left[ \\mathbf{1}_{\\{d_T(x,y)=r\\}} \\right]\n",
+    "$$\n",
+    "\n",
+    "The order of summation is changed, as the sum of expectation values is equal to the expectation value of the sum.\n",
+    "The latter expectation value of the indicator function is exactly equal to the chance $\\mathbb{P}(d_T(x,y)=r)$ for given $x, y$.\n",
+    "For the uniformly distributed $X, Y$, we find $\\mathbb{P}(X = x) = \\frac{1}{V} = \\mathbb{P}(Y = y)$.\n",
+    "This allows us to write the right hand side as follows.\n",
+    "\n",
+    "$$\n",
+    "\\frac{1}{V} \\mathbb{E}\\left[ \\rho_T(r)\\right]\n",
+    "    = \\frac{1}{V^2} \\sum_{x=0}^{V-1}\\sum_{y=0}^{V-1} \\mathbb{P}(d_T(x,y)=r)\n",
+    "    = \\sum_{x=0}^{V-1}\\sum_{y=0}^{V-1} \\mathbb{P}(X = x) \\mathbb{P}(Y = y) \\mathbb{P}(d_T(x,y)=r)\n",
+    "    = \\mathbb{P}(d_T(X,Y)=r),\n",
+    "$$\n",
+    "\n",
+    "which is what we sought.\n",
+    "\n",
+    "Using this result, it is just a matter of writing out the definition of an expectation value to get to the result.\n",
+    "\n",
+    "$$\n",
+    "\\mathbb{E}[d_T(X,Y)] = \\sum_{r=0}^\\infty r\\, \\mathbb{P}(d_T(X,Y) = r) = \\frac{1}{V}\\sum_{r=0}^\\infty r\\, \\mathbb{E}[ \\rho_T(r) ].\n",
+    "$$"
    ]
   },
   {
@@ -475,7 +506,36 @@
     }
    },
    "source": [
-    "YOUR ANSWER HERE"
+    "**To proof**\n",
+    "\n",
+    "$$\n",
+    "\\mathbb{E}[d_T(X,Y)] \\approx c\\,V^{1/d_H}, \\qquad c = \\int_0^\\infty \\mathrm{d}x\\,x\\,f(x)\n",
+    "$$\n",
+    "\n",
+    "**Proof**\n",
+    "\n",
+    "$$\n",
+    "\\mathbb{E} \\left[ d_T(X,Y) \\right]\n",
+    "    = \\frac{1}{V} \\sum_{r=0}^\\infty r\\, \\mathbb{E} \\left[ \\rho_T(r) \\right]\n",
+    "    = \\frac{1}{V} \\sum_{r=0}^\\infty rV^{1-1/d_H}f(rV^{-1/d_H})\n",
+    "    = \\frac{1}{V} \\sum_{r=0}^\\infty xV^{1/d_H} \\cdot V^{1-1/d_H}f(x)\n",
+    "    = \\sum_{r=0}^\\infty xf(x),\n",
+    "$$\n",
+    "where the first equality sign is due to (2), the second due to the given assumption, the third using $x = rV^{-1/d_H}$.\n",
+    "\n",
+    "Now we approximate the summation by an integral.\n",
+    "\n",
+    "$$\n",
+    "\\sum_{r=0}^\\infty xf(x)\n",
+    "    \\approx \\int_{r=0}^\\infty xf(x)dr\n",
+    "    = V^{1/d_H} \\int_{x=0}^\\infty xf(x)dx\n",
+    "    = cV^{1/d_H},\n",
+    "$$\n",
+    "using $\\frac{dr}{dx} = V^{1/d_H}$ for substitution.\n",
+    "This yields the desired approximation\n",
+    "$$\n",
+    " \\mathbb{E} \\left[ d_T(X,Y) \\right] \\approx cV^{1/d_H}.\n",
+    "$$"
    ]
   },
   {
@@ -602,6 +662,7 @@
     "    \n",
     "    for idx_model, model in enumerate(models):\n",
     "        # Calculate mean and standard deviation of the expectations.\n",
+    "        # TODO: Look at whether I store the right data and do the right calculations.\n",
     "        mu = np.mean(expectations[model], 1)\n",
     "        sigma = np.std(expectations[model], 1)\n",
     "\n",