Chapter 10

Stochastic integrals

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Problem 10.1

Prove (10.5) in Proposition 10.3.

Setup: let MM be a square integrable martingale and suppose

Hs(ω)=∑j=1JKj  1(aj,bj](s),(10.3) H_{s}(\omega )=\sum _{j=1}^{J} K_{j} \; \mathbb {1}_{\left(a_{j}, b_{j}\right]}(s), \tag {10.3}

where each KjK_{j} is Faj\mathcal{F}_{a_{j}} measurable and bounded, and the intervals (aj,bj]\left(a_{j}, b_{j}\right] satisfy a1<b1≤a2<b2≤⋯≤aJ<bJa_{1}<b_{1} \leq a_{2}<b_{2} \leq \cdots \leq a_{J}<b_{J}. Define

Nt=∑j=1JKj(Mt∧bj−Mt∧aj).(10.4) N_{t}=\sum _{j=1}^{J} K_{j}\left(M_{t \wedge b_{j}}-M_{t \wedge a_{j}}\right). \tag {10.4}

Proposition 10.3: With HH as in (10.3) and NN defined by (10.4), NtN_{t} is a continuous martingale,

E[N∞2]=E[∫0∞Hs2 d⟨M⟩s], \mathbb {E}\left[N_{\infty }^{2}\right]=\mathbb {E}\left[\int _{0}^{\infty } H_{s}^{2} \, d\langle M\rangle _{s}\right],

and

⟨N⟩t=∫0tHs2 d⟨M⟩s.(10.5) \langle N\rangle _{t}=\int _{0}^{t} H_{s}^{2} \, d\langle M\rangle _{s}. \tag {10.5}
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Problem 10.2

Check the consistency of the first two extensions of the definition of stochastic integrals.

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Problem 10.3

Show that if MM is a continuous square integrable martingale, and TT a finite stopping time, then

∫0∞  1[0,T]dMs=MT \int _{0}^{\infty } \; \mathbb {1}_{[0, T]} d M_{s}=M_{T}
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Problem 10.4

Show that if Nt=∫0tHsdMsN_{t}=\int_{0}^{t} H_{s} d M_{s} where MM is a continuous square integrable martingale, HH is predictable, and E[∫0∞Hs2d⟨M⟩s]<∞\mathbb {E}\left[\int_{0}^{\infty } H_{s}^{2} d\langle M\rangle_{s}\right]<\infty, and Lt=∫0tKsdNsL_{t}=\int_{0}^{t} K_{s} d N_{s}, where KK is predictable and E[∫0∞Ks2d⟨N⟩s]<∞\mathbb {E}\left[\int_{0}^{\infty } K_{s}^{2} d\langle N\rangle_{s}\right]<\infty, then

Lt=∫0tHsKsdMs L_{t}=\int _{0}^{t} H_{s} K_{s} d M_{s}
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Problem 10.5

Show that if M,HM, H, and NN are as in Exercise 10.4, then ⟨M,N⟩t=∫0tHsd⟨M⟩s\langle M, N\rangle_{t}=\int_{0}^{t} H_{s} d\langle M\rangle_{s}.

Hint: Derive a formula for ⟨N+M⟩t\langle N+M\rangle_{t} from the fact that

Nt+Mt=∫0t(1+Hs)dMs N_{t}+M_{t}=\int _{0}^{t}\left(1+H_{s}\right) d M_{s}
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Problem 10.6

Suppose that MM and LL are square integrable martingales, HH is predictable and satisfies (10.2), and Nt=∫0tHsdMsN_{t}=\int_{0}^{t} H_{s} d M_{s}. Show that

⟨N,L⟩t=∫0tHsd⟨M,L⟩s(10.13) \langle N, L\rangle _{t}=\int _{0}^{t} H_{s} d\langle M, L\rangle _{s} \tag {10.13}

Sometimes the stochastic integral of HH with respect to MM is defined to be the square integrable martingale NN for which (10.13) holds for all square integrable martingales LL.

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Problem 10.7

Show that if MM and NN are square integrable martingales with continuous paths, then

⟨M,N⟩t≤(⟨M⟩t)1/2(⟨N⟩t)1/2 \langle M, N\rangle _{t} \leq \left(\langle M\rangle _{t}\right)^{1 / 2}\left(\langle N\rangle _{t}\right)^{1 / 2}

Hint: Imitate an appropriate proof of the Cauchy-Schwarz inequality. This result is a special case of the inequality of Kunita-Watanabe.

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