4  Conditional expectation

For part i, there is several ways to do this, here are two of them:

• a mild variation of Equation 3.45, using Jensen’s inequality (cf. Theorem 3.24 iv): observe that the function \(f:\R \to \R\) given by \(f(x)=x^2\) is convex since \(f''(x)=2 \geq 0\). Now apply Jensen’s inequality to the random variable \(\lvert X \rvert\): \[\big( \E[\lvert X \rvert] \big)^2 \leq \E \left[ \lvert X \rvert^2 \right]=\E \left[ X^2 \right]<\infty\] and hence also \(\E[ \lvert X \rvert]<\infty\).
• alternatively you could use that if we define the functions \(f(x)=\lvert x \rvert\) and \(g(x)=x^2+1\), then you can readily check that \(f(x) \leq g(x)\) for all \(x \in \R\) and hence in terms of random variables, \(f(X) \leq g(X)\) (pointwise). Applying the non-negativity property (cf. Theorem 3.24 ii, in particular Equation 3.43) then yields \[\E[\lvert X \rvert] \leq \E[X^2+1],\] and by the linearity property (cf. Theorem 3.24 i) plus Exercise 3.5, \[\E[X^2+1]=\E[X^2]+\E[1]=\E[X^2]+1<\infty.\]

For part ii, recall from Definition 3.22 (see the three cases) that in general, \(\E[X]\) may not exist, or be equal to \(\pm \infty\), or be a real number. Since we have that \(\E[ \lvert X \rvert]<\infty\), we are in the first case in which it is guaranteed that \(\E[X] \in \R\).

For part iii, working out the brackets and using the linearity property (cf. Theorem 3.24 i) yields \[\begin{align*} \E[(X-a)^2] &= \E \left[ X^2 -2aX+a^2 \right] \\ &= \E \left[ X^2 \right] -2a\E[X]+a^2 \end{align*} \] and now just plug in the results from parts i & ii.

For part iv, as part iii shows, we need to find the minimiser of the function \[f(a)=\E \left[ X^2 \right] -2a\E[X]+a^2.\] Observe that this is simply a second order polynomial, and hence differentiating it or throwing the well known formula at it, you can readily check that the minimiser is \(a=\E[X]\).

If needs be, recall the definition of an indicator function from Subsection 2.4.2.

For part i, we have that \[\begin{aligned}\mathbf{1}_A(\omega) \mathbf{1}_B(\omega)=1 &\iff \mathbf{1}_A(\omega)=1 \text{ and } \mathbf{1}_B(\omega)=1 \\ &\iff \omega \in A \text{ and } \omega \in B \\ &\iff \omega \in A \cap B \\ &\iff \mathbf{1}_{A \cap B}(\omega)=1, \end{aligned} \] and since both sides are either \(0\) or \(1\) it also follows that \[\mathbf{1}_A(\omega) \mathbf{1}_B(\omega)=0 \iff \mathbf{1}_{A \cap B}(\omega)=0.\]

For part ii, similarly \[\begin{aligned}\mathbf{1}_A(\omega) + \mathbf{1}_B(\omega)=0 &\iff \mathbf{1}_A(\omega)=0 \text{ and } \mathbf{1}_B(\omega)=0 \\ &\iff \omega \not\in A \text{ and } \omega \not\in B \\ &\iff \omega \not\in A \cup B \\ &\iff \mathbf{1}_{A \cup B}(\omega)=0. \end{aligned} \] Since \(A\) and \(B\) are disjoint, \(\mathbf{1}_A + \mathbf{1}_B\) only takes the values \(0\) and \(1\) and we’re again done.

If \(A\) and \(B\) are not disjoint i.e. \(A \cap B \not= \emptyset\), then for any \(\omega \in A \cap B\) we have \[\mathbf{1}_A(\omega) + \mathbf{1}_B(\omega)=2 \not= 1=\mathbf{1}_{A \cup C}(\omega).\]

Let’s do the expectation first. Looking at Equation 4.2 in Definition 4.5, since \(\P(C)=1\) (cf. Definition 1.55) we’ll need to show that \(\E[\mathbf{1}_C X]=\E[X]\). But observe that for any \(\omega \in C\), by def of the indicator function (cf. Subsection 2.4.2) \[\mathbf{1}_C(\omega) X(\omega)=1 \cdot X(\omega)=X(\omega)\] i.e. \(\mathbf{1}_C X=X\) on \(C\), and since \(C\) is an almost sure event, \(\mathbf{1}_C X=X\) a.s. (recall from Subsection 2.3.1). And then it follows from Equation 3.44 in Theorem 3.24 that \(\E[\mathbf{1}_C X]=\E[X]\).

For the probability, just use that \(\P(A)=\E[\mathbf{1}_A]\) (cf. Equation 2.21) and \(\P(A \, | \, C)=\E[\mathbf{1}_A \, | \, C]\) (as noted in Definition 4.5) to make it follow immediately from the above, or otherwise do it separately if you prefer of course, for instance by breaking \(A\) up in a part inside \(C\) and a part outside \(C\) i.e.  \[A=(A \cap C) \cup (A \cap C^c)\] and taking it from there.

For the Law of Total Expectation, first plugging in the def (cf. Definition 4.5) we see that \[\sum_{i=1}^n \E[X \, | \, C_i] \P(C_i)=\sum_{i=1}^n \E[\mathbf{1}_{C_i} X].\] Using linearity (cf. Theorem 3.24 i) we can bring the summation inside the expectation and then it remains to make the observation that for any \(\omega \in \Omega\) \[\sum_{i=1}^n \mathbf{1}_{C_i}(\omega)=1.\] For this, since \(C_1, \ldots, C_n\) is partition of \(\Omega\), any \(\omega \in \Omega\) is an element of exactly one \(C_i\), so that \(\mathbf{1}_{C_i}(\omega)=1\) and \(\mathbf{1}_{C_j}(\omega)=0\) for \(j \not= i\). See also Exercise 4.2 ii.

For the Law of Total Probability, just take Equation 4.24 and plug in \(X=\mathbf{1}_A\).

There’s several ways to approach this, of which one is to use the Law of Total Probability from Exercise 4.4 (that’s why the question is included here of course, just to see a cute application). Denoting by \(A\) the event that Kevin catches more than 12 squirrels next week, and by \(C_i\) the event that he gets to go for \(i\) walks next week, for \(i=0,\ldots,7\), these \(C_i\)’s form a partition of the outcome space (which you can formulate in any (sensible) way you want). Hence by the law of total probability \[\P(A)=\sum_{i=0}^7 \P(A \, | \, C_i) \P(C_i).\] Now, if Kevin gets to go out for six days or fewer then he can’t possibly catch more than 12 squirrels, so \(\P(A \, | \, C_i)=0\) for \(i=1,\ldots,6\). Further, conditional on \(C_7\), the number of squirrels that he will catch (assuming independence between walks) is 14 if he catches two every day i.e. with prob \(0.3^7\) and it is 13 if he catches two on six days and one on the remaining day i.e. with prob \(7 \cdot 0.3^6 \cdot 0.6\). So in conclusion \[\P(A)=\P(A \, | \, C_7) \P(C_7)=\big( 0.3^7 + 7 \cdot 0.3^6 \cdot 0.6 \big) \cdot 0.8^7 \approx 0.0007.\] So not the best odds for our poor friend…

For part i, from Definition 4.5 we know that \[\E[X \, | \, A]=\frac{\E[\mathbf{1}_{A} X]}{\P(A)}.\] Since \[\E[\mathbf{1}_{A} X]=\E[\mathbf{1}_{A} c]=c \E[\mathbf{1}_{A}]=c \P(A)\] (by linearity, cf. Theorem 3.24 i, and Equation 2.21), so indeed \(\E[X \, | \, A]=c\).

For part ii, note that we need to show that the constant random variable \(Z:=c\) is integrable (but that’s trivial) and satisfies conditions i & ii of Definition 4.9 to confirm that is indeed (a version of) \(\E[X \, | \, \mathcal{G}]\). Both are trivially true: a constant function is measurable with respect to any \(\sigma\)-algebra (cf. Exercise 2.3 e.g.), and for any event \(A \in \mathcal{G}\), \(\E[\mathbf{1}_A Z]=\E[\mathbf{1}_A X]\) is obviously true as both \(X\) and \(Z\) are constantly equal to \(c\).

For part i, obviously we need to appeal to Definition 4.9. Since \(\mathcal{G}\) is finite, we can in particular make use of Equation 4.10. First observe that the atoms of \(\mathcal{G}\) (i.e. ignoring \(\emptyset\), the “smallest” elements that form a partition of \(\Omega\) — cf. Proposition 1.38) are \[A_1:=\{\omega_2,\omega_3\} \quad \text{and} \quad A_2:=\{\omega_1,\omega_4,\omega_5\}.\] Now we just need to work out \(\E[X \, | \, A_1]\) and \(\E[X \, | \, A_2]\). From Definition 4.5 we get the expression \[\E[X \, | \, A_i]=\frac{\E[\mathbf{1}_{A_i} X]}{\P(A_i)}. \tag{4.25}\] Using countable additivity (cf. Definition 1.50) we can easily compute \(\P(A_1)=\P(\{\omega_2\})+\P(\{\omega_3\})=2/5\) and similarly \(\P(A_2)=3/5\). Further don’t get confused: \(\mathbf{1}_{A_1} X\) is “just” a random variable that is a product of two random variables: the indicator function \(\mathbf{1}_{A_1}\) and \(X\), so you can easily work out what mapping it is here: \[\mathbf{1}_{A_1}(\omega) X(\omega)= \begin{cases} 2 & \text{if $\omega=\omega_2$} \\ 3 & \text{if $\omega=\omega_3$} \\ 0 & \text{if $\omega \in A_2=\{\omega_1,\omega_4,\omega_5\}$.} \end{cases} \] Hence this is a random variable with range \(\{0,2,3\}\) i.e. a discrete one (cf. Subsection 2.4.1), and using the standard expression for its expectation (cf. Proposition 3.23 with \(h(x)=x\)) we find \[\begin{aligned} \E[\mathbf{1}_{A_1} X] &= 2 \cdot \P(\mathbf{1}_{A_1} X=2) + 3 \cdot \P(\mathbf{1}_{A_1} X=3) + 0 \cdot \P(\mathbf{1}_{A_1} X=0) \\ &= 2 \cdot \P(\{\omega_2\}) + 3 \cdot \P(\{\omega_3\})=1 \end{aligned} \] and hence Equation 4.25 yields \[\E[X \, | \, A_1]=\frac{1}{2/5}=\frac{5}{2}.\] In exactly the same way you can work out that \(\E[X \, | \, A_2]=10/3\). Finally, going back to Equation 4.10 we can now conclude that \[\E[X \, | \, \mathcal{G}](\omega)=\begin{cases} 5/2 & \text{if $\omega \in A_1=\{\omega_2,\omega_3\}$} \\ 10/3 & \text{if $\omega \in A_2=\{\omega_1,\omega_4,\omega_5\}$.} \end{cases} \tag{4.26}\]

For part ii, repeatedly using the standard expression for the expectation of a discrete random variable (cf. Proposition 3.23) we can work out \[\begin{align*} \E[X] &= 1 \cdot \P(X=1) + 2 \cdot \P(X=2) + \ldots + 5 \cdot \P(X=5) \\ &= 1 \cdot \P(\{\omega_1\}) + 2 \cdot \P(\{\omega_2\}) + \ldots + 5 \cdot \P(\{\omega_5\}) &=\frac{15}{5}=3 \end{align*} \] and, using Equation 4.26, \[\begin{align*} \E \big[ \E[X \, | \, \mathcal{G}] \big] &= \frac{5}{2} \cdot \P(A_1) + \frac{10}{3} \cdot \P(A_2) \\ &= \frac{5}{2} \cdot \frac{2}{5} + \frac{10}{3} \cdot \frac{3}{5}=3. \end{align*} \] So they’re indeed equal, as Theorem 4.15 i dictates they should be!

For part i, of course one possible way to verify a pointwise identity is to take an arbitrary \(\omega \in \Omega\) and plug it into the random variables involved, i.e. \(Z\), \(\widetilde{Z}\), \(\mathbf{1}_{\{Z<\widetilde{Z}\}}\) and \(\mathbf{1}_{\{Z>\widetilde{Z}\}}\), and convince yourself that the resulting identity involving real numbers holds (where you may have to distinguish a few cases etc.).

Alternatively, observe that the events \(\{Z \leq \widetilde{Z}\}\) and \(\{Z>\widetilde{Z}\}\) form a partition of \(\Omega\) i.e. that (pointwise) \[\mathbf{1}_{\{Z \leq \widetilde{Z}\}}+\mathbf{1}_{\{Z>\widetilde{Z}\}}=1\] (see also Exercise 4.2 ii). This yields \[\begin{align*} \left\lvert Z-\widetilde{Z} \right\rvert &= \left( \mathbf{1}_{\{Z \leq \widetilde{Z}\}}+\mathbf{1}_{\{Z>\widetilde{Z}\}} \right) \left\lvert Z-\widetilde{Z} \right\rvert \\ &= \mathbf{1}_{\{Z \leq \widetilde{Z}\}} \left\lvert Z-\widetilde{Z} \right\rvert + \mathbf{1}_{\{Z>\widetilde{Z}\}} \left\lvert Z-\widetilde{Z} \right\rvert \\ &= \mathbf{1}_{\{Z \leq \widetilde{Z}\}} \left( \widetilde{Z}-Z \right)+\mathbf{1}_{\{Z>\widetilde{Z}\}} \left( Z-\widetilde{Z} \right), \end{align*} \] where the final equation uses for the first term, \(\mathbf{1}_{\{Z \leq \widetilde{Z}\}}=1\) if and only if we are on the event \(\{Z \leq \widetilde{Z}\}\) where \(\lvert Z-\widetilde{Z} \rvert=\widetilde{Z}-Z\), and analogue for the second term. (Of course, also on any of the substeps here where you prefer to be more explicit to see better what’s going on, plug in an \(\omega \in \Omega\)!)

For part ii, since \(Z\) and \(\widetilde{Z}\) are both versions of \(\E[X \, | \, \mathcal{G}]\), we know from Definition 4.9 that they are both integrable, \(\mathcal{G}\)-measurable and \[\E[\mathbf{1}_A Z]=\E[\mathbf{1}_A X]=\E \left[ \mathbf{1}_A \widetilde{Z} \right] \quad \text{for all } A \in \mathcal{G}. \tag{4.27}\]

Now, first note that \(\{Z \leq \widetilde{Z}\} \in \mathcal{G}\) and \(\{Z>\widetilde{Z}\} \in \mathcal{G}\). The nicest argument for this is to write \[\{Z \leq \widetilde{Z}\}=\{Z -\widetilde{Z} \leq 0\},\] which is the inverse image of the Borel set \((-\infty,0]\) under \(Z -\widetilde{Z}\). Since \(Z\) and \(\widetilde{Z}\) are both \(\mathcal{G}\)-measurable so is \(Z -\widetilde{Z}\) (cf. Proposition 2.11 iv) and hence that inverse image is indeed an element of \(\mathcal{G}\) (by def, recall from Definition 2.6). Analogue for \(\{Z>\widetilde{Z}\}\) (or use that it’s the complement of course, and that \(\sigma\)-algebras are closed under complements).

This means that we can use these inverse images as our choice for \(A\) in Equation 4.27. This yields that \[\E \left[ \mathbf{1}_{\{Z \leq \widetilde{Z}\}} \widetilde{Z} \right]=\E \left[ \mathbf{1}_{\{Z \leq \widetilde{Z}\}} Z \right]\] and hence by linearity (cf. Theorem 3.24 i) \[\E \left[ \mathbf{1}_{\{Z \leq \widetilde{Z}\}} \left( \widetilde{Z}-Z \right) \right]=0.\] Similarly it follows that \[\E \left[ \mathbf{1}_{\{Z>\widetilde{Z}\}} \left( Z-\widetilde{Z} \right) \right]=0,\] and hence by part i and linearity again also \[\E \left[ \left\lvert Z-\widetilde{Z} \right\rvert \right]=0.\]

For part iii, given the result in part ii, all we need to do is appeal to almost sure equivalence, cf. Theorem 3.24 iii!

For property ii, following Definition 4.9, as in Exercise 4.6 ii for instance, what we need to do is show that if we define \(Z:=X\), then \(Z\) satisfies all conditions in Definition 4.9. But that’s all obvious: \(Z\) is integrable and also \(\mathcal{G}\)-measurable since \(X\) is, and trivially also \(\E[\mathbf{1}_A X]=\E[\mathbf{1}_A Z]\) for all \(A \in \mathcal{G}\) of course!

For property iii, pursuing the same strategy as for property ii, define the constant random variable \(Z:=\E[X]\). Then \(Z\) is trivially integrable, measurable with respect to any \(\sigma\)-algebra (cf. Exercise 2.3) hence also with respect to \(\mathcal{G}\), and finally we need to show that it holds that \[\E[\mathbf{1}_A Z]=\E[\mathbf{1}_A X] \quad \text{for all } A \in \mathcal{G}.\] Note that on the left hand side, since \(Z=\E[X]\) is a constant, we can use linearity (cf. Theorem 3.24 i) together with Equation 2.21 to write \[\E[\mathbf{1}_A Z]=\E \big[ \mathbf{1}_A \E[X] \big]=\E[X] \E[\mathbf{1}_A]=\E[X] \P(A)\] so that the equation we need to show can be rewritten to \[\E[X] \P(A)=\E[\mathbf{1}_A X] \quad \text{for all } A \in \mathcal{G}. \tag{4.28}\] Now we need to start making use of the assumption that \(\mathcal{G}\) is trivial, i.e. that for any \(A \in \mathcal{G}\) we have either that \(\P(A)=0\) or \(\P(A)=1\).

First assume that \(\P(A)=0\). Then clearly the left hand side of Equation 4.28 equals \(0\). For the right hand side, observe that \(\mathbf{1}_A X=0\) a.s. (because by assumption \(A^c\) is an almost sure event, cf. Definition 1.55 & Exercise 1.12 i, and for any \(\omega \in A^c\) we have \(\mathbf{1}_A(\omega) X(\omega)=0 \cdot X(\omega)=0\)) so that we get from Equation 3.44 in Theorem 3.24 that \(\E[\mathbf{1}_A X]=\E[0]=0\) (the latter by Exercise 3.5).

Now assume that \(\P(A)=1\). Now the left hand side of Equation 4.28 equals \(\E[X]\), and by the same arguments as just used we now see that \(\mathbf{1}_A X=X\) a.s. and hence also \(\E[\mathbf{1}_A X]=\E[X]\).

For property vi, following the same arguments as we used for property iii above, we end up again needing to show that \[\E[X] \P(A)=\E[\mathbf{1}_A X] \quad \text{for all } A \in \mathcal{G}. \tag{4.29}\]

For this, fix any \(A \in \mathcal{G}\). Recall that we discussed independence in Section 2.6. We know that since \(A \in \mathcal{G}\), \(\mathbf{1}_A\) is \(\mathcal{G}\)-measurable (cf. Subsection 2.4.2). Because \(\sigma(\mathbf{1}_A)\), the \(\sigma\)-algebra generated by \(\mathbf{1}_A\), is the smallest \(\sigma\)-algebra that makes \(\mathbf{1}_A\) measurable we also have that \(\sigma(\mathbf{1}_A) \subseteq \mathcal{G}\) (cf. Proposition 2.9). Now, by assumption the \(\sigma\)-algebras \(\sigma(X)\) and \(\mathcal{G}\) are independent. Looking at the def, cf. Definition 2.30, it readily follows that \(\sigma(X)\) and \(\sigma(\mathbf{1}_A)\) are independent as well. And hence by Proposition 2.31, \(X\) and \(\mathbf{1}_A\) are independent random variables. This then implies, using e.g. Theorem 3.24 viii, that \[\E[\mathbf{1}_A X]=\E[\mathbf{1}_A] \E[X].\] Now just plug in Equation 2.21 to indeed arrive at Equation 4.29.

For part i, following the instructions to work out the brackets we get that \[\begin{aligned} \E[(X+Y)^2 \, | \, \mathcal{G}] &= \E[X^2+2XY+Y^2 \, | \, \mathcal{G}] \\ &=\E[X^2 \, | \, \mathcal{G}]+2 \E[XY \, | \, \mathcal{G}]+\E[Y^2 \, | \, \mathcal{G}] \\ &=\E[X^2 \, | \, \mathcal{G}]+2Y \E[X \, | \, \mathcal{G}]+\E[Y^2 \, | \, \mathcal{G}] \\ &=\E[X^2]+2Y \E[X]+Y^2 \\ &=\frac{91}{6}+7Y+Y^2 \end{aligned} \] (observe: it is a random variable, as we know such a conditional expectation generally is!), where the second equality uses linearity (Theorem 4.14 i), the third by “taking out” \(Y\) (indeed \(Y\) is \(\mathcal{G}=\sigma(Y)\)-measurable, cf. Proposition 2.9, so we can use Theorem 4.15 v), and the fourth is by Theorem 4.14 ii & vi.

For part ii, with \(\Phi(x,y)=(x+y)^2\) we can compute for any \(y \in \R\) \[\varphi(y)=\E[\Phi(X,y)]=\E[(X+y)^2]=\E[X^2]+2y \E[X]+y^2=\frac{91}{6}+7y+y^2\] (we just work out the brackets and use linearity properties of \(\E\), cf. Theorem 3.24 i) and hence by Theorem 4.15 vii \[\E[(X+Y)^2 \, | \, \mathcal{G}]=\E[\Phi(X,Y)]=\varphi(Y)=\frac{91}{6}+7Y+Y^2\] (reassuringly the same result as part i!).

For part iii, using Theorem 4.14 i, the result from i/ii, and linearity of \(\E\) (cf. Theorem 3.24 i) we can compute \[\E[(X+Y)^2]=\E \big[ \E[(X+Y)^2 \, | \, \mathcal{G}] \big] = \E \left[ \frac{91}{6}+7Y+Y^2 \right] = \frac{91}{6} + 7 \cdot \frac{7}{2} + \frac{91}{6}=\frac{329}{6}\]

If we take for \(\mathcal{H}\) a “trivial” \(\sigma\)-algebra, e.g. \(\mathcal{H}=\{\emptyset,\Omega\}\) in the Tower Property \[\E \big[ \E[X \, | \, \mathcal{G}] \, \big| \, \mathcal{H} \big]=\E[X \, | \, \mathcal{H}]\] then using Theorem 4.15 iii it simplifies to \[\E \big[ \E[X \, | \, \mathcal{G}] \big]=\E[X].\]

For part i, first note that since \(X^2\) is an integrable random variable, its conditional expectation \(\E[X^2 \, | \, \mathcal{G}]\) is well defined. From Theorem 4.15 i we get that \[\E \left[ \E[X^2 \, | \, \mathcal{G}] \right]=\E[X^2]<\infty.\] Jensen’s inequality (cf. Theorem 4.14 iv) with \(f(x)=x^2\) yields \[\left( \E[X \, | \, \mathcal{G}] \right)^2 \leq \E[X^2 \, | \, \mathcal{G}] \quad \text{a.s.}\] and hence by non-negativity of \(\E\) \[\E \left[ \left( \E[X \, | \, \mathcal{G}] \right)^2 \right] \leq \E \left[ \E[X^2 \, | \, \mathcal{G}] \right]<\infty.\]

For part ii, first assume that \(\E[(X-Y)G]=0\) for all \(G \in \mathbf{L}\). Then for any \(G \in \mathbf{L}\) \[\begin{aligned} \E[(X-G)^2] &= \E[(X-Y+Y-G)^2] \\ &= \E[(X-Y)^2] + 2 \E[(X-Y)(Y-G)] + \E[(Y-G)^2] \\ &= \E[(X-Y)^2] + \E[(Y-G)^2] \\ &\geq \E[(X-Y)^2], \end{aligned} \] where the third equality uses that since \(Y,G \in \mathbf{L}\), also \(Y-G \in \mathbf{L}\) (second moment condition is trivial to check) and hence \(\E[(X-Y)(Y-G)]=0\) on account of our assumption.

Now assume that \(\E[(X-Y)^2] \leq \E[(X-G)^2]\) for all \(G \in \mathbf{L}\). Suppose that a \(G \in \mathbf{L}\) exists so that \(\E[(X-Y)G]\not=0\) and let’s derive a contradiction. Without loss of generality, \(\E[(X-Y)G]>0\) (otherwise replace \(G\) by \(-G\)). Further without loss of generality, \(\E[G^2]=1\) (otherwise, scale \(G\) by an appropriate constant factor — this would only not be possible if \(\E[G^2]=0\), but that would imply \(G=0\) a.s. and hence \(\E[(X-Y)G]=0\) (both by almost sure equivalence, cf. Theorem 3.24 iii)). Now, set \(\alpha:=\E[(X-Y)G]>0\) and define \(H=Y+\alpha G \in \mathbf{L}\). Then \[\begin{aligned} \E[(X-H)^2] &= \E[(X-Y+Y-H)^2] \\ &= \E[(X-Y)^2] + 2 \E[(X-Y)(Y-H)] + \E[(Y-H)^2] \\ &= \E[(X-Y)^2] -2\alpha \E[(X-Y)G] + \alpha^2 \E[G^2] \\ &= \E[(X-Y)^2] -2\alpha \cdot \alpha + \alpha^2 \\ &= \E[(X-Y)^2] - \alpha^2 \\ &< \E[(X-Y)^2], \end{aligned} \] violating our assumption.

For part iii, first note that if \(Z \in \mathbf{L}\) is any minimiser (assuming one exists), then by part ii it satisfies \(\E[(X-Z)G]=0\) for all \(G \in \mathbf{L}\). For any \(A \in \mathcal{G}\), using this with \(G=\mathbf{1}_A\) yields \[0=\E[(X-Z)\mathbf{1}_A]=\E[\mathbf{1}_A X]-\E[\mathbf{1}_A Z],\] and hence by Definition 4.9 \(Z\) is version of \(\E[X \, | \, \mathcal{G}]\). Since all versions are equal a.s., there is also at most one (up to a.s. equivalence) minimiser.

Now for the hard part: writing \(Z=\E[X \, | \, \mathcal{G}]\), showing that \(Z\) is a minimiser. Using ii it is enough to show that \(\E[(X-Z)G]=0\) for all \(G \in \mathbf{L}\). Again by def (cf. Definition 4.9) we know for any \(A \in \mathcal{G}\) that \[\E[(X-Z)\mathbf{1}_A]=\E[\mathbf{1}_A X]-\E[\mathbf{1}_A Z]=0, \tag{4.30}\] so we need to worry about how to extend this from all indicator functions to all random variables (both in \(\mathbf{L}\)). You can do this in a constructive way: fix some \(G \in \mathbf{L}\), without loss of generality assume \(G \geq 0\) (otherwise just treat its positive and negative parts separately). Now approximate \(G\) (pointwise) by a non-decreasing sequence of non-negative \(\mathcal{G}\)-measurable step functions, say \(G^{(1)} \leq G^{(2)} \leq \ldots\), e.g. as in Subsection 3.4.2, and by writing each \(G^{(n)}\) as a sum of indicator functions we get that \(\E[(X-Z)G^{(n)}]=0\) from Equation 4.30. So if can show that \[\lim_{n \to \infty} \E[(X-Z)G^{(n)}]=\E[(X-Z)G] \tag{4.31}\] then we’re done. For this, note that (pointwise) \[\lvert (X-Z)G^{(n)} \rvert \leq \lvert (X-Z) \rvert \lvert G^{(n)} \rvert \leq \lvert (X-Z) \rvert G\] and the ultimate right hand side is integrable (because they all of \(X\), \(Z\) and \(G\) have finite second moments, use e.g. Hölder’s inequality with \(p=q=2\), cf. Theorem 3.24 vii), and so Equation 4.31 follows from the DCT (cf. Theorem 3.27). Note: this may well be a sledgehammer to crack a nut — it’s too late at night and I can’t seem to think of an easier/more elegant solution (without using facts about \(L^2\) spaces that we haven’t discussed at least), I’d be keen to hear it if you see one though!

For part iv, first note that \[\E \big[ X-\E[X \, | \, \mathcal{G}] \, | \, \mathcal{G} \big] = \E[X \, | \, \mathcal{G}] - \E \big[ \E[X \, | \, \mathcal{G}] \, | \, \mathcal{G} \big] = \E[X \, | \, \mathcal{G}]-\E[X \, | \, \mathcal{G}]=0 \quad \text{a.s.}, \tag{4.32}\] where the first equality uses linearity (cf. Theorem 4.14 i) and the second Theorem 4.15 ii since \(\E[X \, | \, \mathcal{G}]\) is \(\mathcal{G}\)-meas by definition. Hence also \[ \begin{aligned} & \E \big[ (X-\E[X \, | \, \mathcal{G}]) (\E[X \, | \, \mathcal{G}]-\E[X \, | \, \mathcal{H}]) \, | \, \mathcal{G} \big] \\ & = (\E[X \, | \, \mathcal{G}]-\E[X \, | \, \mathcal{H}]) \E \big[ X-\E[X \, | \, \mathcal{G}] \, | \, \mathcal{G} \big] =0 \end{aligned} \tag{4.33}\] where the first equality uses that \(\E[X \, | \, \mathcal{G}]-\E[X \, | \, \mathcal{H}]\) is \(\mathcal{G}\)-meas (the first term is that by definition, the second term is by def \(\mathcal{H}\)-meas and \(\mathcal{H} \subseteq \mathcal{G}\)) together with Theorem 4.15 ii. Taking expectations on both sides of Equation 4.33 yields, also using Theorem 4.15 i \[\E \big[ (X-\E[X \, | \, \mathcal{G}]) (\E[X \, | \, \mathcal{G}]-\E[X \, | \, \mathcal{H}]) \big]=0. \tag{4.34}\] Here is why this is useful. We can write (using the same trick as in part ii) \[ \begin{aligned} \E \big[ (X-\E[X \, | \, \mathcal{H}])^2 \big] &= \E \big[ (X-\E[X \, | \, \mathcal{G}]+\E[X \, | \, \mathcal{G}]-\E[X \, | \, \mathcal{H}])^2 \big] \\ &= \E \big[ (X-\E[X \, | \, \mathcal{G}])^2 \big]\\ & \qquad +2 \E \big[ (X-\E[X \, | \, \mathcal{G}]) (\E[X \, | \, \mathcal{G}]-\E[X \, | \, \mathcal{H}]) \big] \\ & \qquad + \E \big[ \big( \E[X \, | \, \mathcal{G}]-\E[X \, | \, \mathcal{H}] \big)^2 \big] \\ &= \E \big[ (X-\E[X \, | \, \mathcal{G}])^2 \big] + \E \big[ \big( \E[X \, | \, \mathcal{G}]-\E[X \, | \, \mathcal{H}] \big)^2 \big] \\ &\geq \E \big[ (X-\E[X \, | \, \mathcal{G}])^2 \big] \end{aligned} \] so that we have arrived at what we wanted to prove. The second equality just works out the brackets in the same way as in part ii, the third equality uses Equation 4.34 and finally the inequality uses non-negativity.