Conditional Mean and Variance for Random Step Integrals

Why Condition One Interval at a Time?

An adapted step integrand is random, but on each time interval its value is fixed by the information available at the interval's left endpoint. The Brownian increment over that interval is independent of that information. This gives a direct way to compute conditional means and variances without treating the entire integral as though its integrand were deterministic.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$ relative to the filtration: for $s<t$, $B_t-B_s$ is independent of $\mathcal{F}_s$ and has distribution $N(0,t-s)$. Fix a finite grid $0=t_0<t_1<\cdots<t_n=T$. Let $H_i$ be an $\mathcal{F}_{t_i}$-measurable random variable with $\mathbb{E}[H_i^2]<\infty$, and set $H_s=H_i$ on $[t_i,t_{i+1})$. This is a progressively measurable step process with $\mathbb{E}\int_0^T H_s^2\,ds<\infty$, so it belongs to $\mathcal{L}^2$. Its Ito integral is the finite sum

M_T=\int_0^T H_s\,dB_s =\sum_{i=0}^{n-1}H_i(B_{t_{i+1}}-B_{t_i}).

For a square-integrable random variable $X$, its conditional variance given $\mathcal{G}$ is $\operatorname{Var}(X\mid\mathcal{G})=\mathbb{E}[(X-\mathbb{E}[X\mid\mathcal{G}])^2\mid\mathcal{G}]$. In the formulas below, conditioning on the left endpoint's information makes the coefficient $H_i$ known. Conditioning earlier may not.

One Interval: Conditional Mean and Variance

Proposition: Let $H$ be $\mathcal{F}_a$-measurable with $\mathbb{E}[H^2]<\infty$, and let $a<b$. For the Ito integral $Y=\int_a^b H\,dB=H(B_b-B_a)$, $$ \mathbb{E}[Y\mid\mathcal{F}_a]=0,\qquad \operatorname{Var}(Y\mid\mathcal{F}_a)=H^2(b-a). $$ The integrand on $[a,b]$ is progressively measurable and square-integrable since $\mathbb{E}\int_a^b H^2\,ds=(b-a)\mathbb{E}[H^2]<\infty$.

Proof. Write $\Delta B=B_b-B_a$ and $h=b-a$. By Brownian increment independence, $\Delta B$ is independent of $\mathcal{F}_a$, with $\mathbb{E}[\Delta B]=0$ and $\mathbb{E}[(\Delta B)^2]=h$. Since $H$ is $\mathcal{F}_a$-measurable, conditional expectation gives $\mathbb{E}[H\Delta B\mid\mathcal{F}_a]=H\mathbb{E}[\Delta B\mid\mathcal{F}_a]=0$. Thus the conditional variance is $\mathbb{E}[H^2(\Delta B)^2\mid\mathcal{F}_a]=H^2h$. The products are integrable because $\mathbb{E}[H^2(\Delta B)^2]=h\mathbb{E}[H^2]<\infty$. $\square$

The coefficient can be random; it is simply known at the instant conditioning begins. In contrast, if we condition on information from before the coefficient is chosen, its square must still be averaged conditionally.

Worked Example: A Brownian Value as the Next Coefficient

Take the grid $0<1<2$, with $H_0=1$ and $H_1=B_1$. Thus $H_s=1$ on $[0,1)$ and $H_s=B_1$ on $[1,2)$. Both coefficients are square-integrable: $\mathbb{E}[1^2]=1$ and $\mathbb{E}[B_1^2]=1$. Compute the terminal integral's conditional mean and variance given $\mathcal{F}_1$, and then its unconditional mean and variance.

The integral is $M_2=B_1+B_1(B_2-B_1)$. Given $\mathcal{F}_1$, $B_1$ is known and $B_2-B_1$ is independent with mean $0$ and variance $1$. Therefore $\mathbb{E}[M_2\mid\mathcal{F}_1]=B_1$, and $\operatorname{Var}(M_2\mid\mathcal{F}_1)=B_1^2$.

Taking expectations gives $\mathbb{E}[M_2]=\mathbb{E}[B_1]=0$. By conditioning, $\mathbb{E}[M_2^2\mid\mathcal{F}_1]=B_1^2+B_1^2=2B_1^2$, so $\operatorname{Var}(M_2)=\mathbb{E}[M_2^2]=2\mathbb{E}[B_1^2]=2$.

Check: the Ito isometry gives $\mathbb{E}[M_2^2]=\mathbb{E}\int_0^2H_s^2\,ds=1+\mathbb{E}[B_1^2]=2$, agreeing with the conditional calculation.

Several Future Intervals

For several steps, individual future increments have conditional mean zero, but their coefficients may depend on information revealed between the conditioning time and those increments. Consequently, the conditional variance of the whole future sum involves the conditional expectation of its future energy, not necessarily the energy itself.

Theorem: Use the finite grid and square-integrable adapted coefficients defined above, and write $\Delta B_i=B_{t_{i+1}}-B_{t_i}$. For $k\le n$, let $M_k=\sum_{i=0}^{k-1}H_i\Delta B_i$, with $M_0=0$. Then $$ \mathbb{E}[M_n-M_k\mid\mathcal{F}_{t_k}]=0,\qquad \operatorname{Var}(M_n\mid\mathcal{F}_{t_k}) =\mathbb{E}\left[\sum_{i=k}^{n-1}H_i^2(t_{i+1}-t_i)\,\middle|\,\mathcal{F}_{t_k}\right]. $$

Proof. Put $U_i=H_i\Delta B_i$ and $h_i=t_{i+1}-t_i$. The one-interval proposition gives $\mathbb{E}[U_i\mid\mathcal{F}_{t_i}]=0$ and $\mathbb{E}[U_i^2\mid\mathcal{F}_{t_i}]=H_i^2h_i$. For $i\ge k$, the tower property implies $\mathbb{E}[U_i\mid\mathcal{F}_{t_k}]=0$, so the conditional mean of the finite sum is zero. If $k\le i<j<n$, then $U_i$ is $\mathcal{F}_{t_j}$-measurable. Hence $\mathbb{E}[U_iU_j\mid\mathcal{F}_{t_j}]=U_i\mathbb{E}[U_j\mid\mathcal{F}_{t_j}]=0$, and conditioning again gives $\mathbb{E}[U_iU_j\mid\mathcal{F}_{t_k}]=0$. All these products are integrable by Cauchy–Schwarz, since each $U_i$ is square-integrable. Expanding the square of the finite sum therefore leaves only diagonal terms. For each $i\ge k$, conditioning first on $\mathcal{F}_{t_i}$ and then on $\mathcal{F}_{t_k}$ gives $\mathbb{E}[U_i^2\mid\mathcal{F}_{t_k}]=\mathbb{E}[H_i^2h_i\mid\mathcal{F}_{t_k}]$. The conditional mean is zero, so this sum of diagonal terms is the stated conditional variance. $\square$

This calculation also verifies the martingale property for the step-integral process at grid times: for $k\le m$, the same conditional-mean argument gives $\mathbb{E}[M_m\mid\mathcal{F}_{t_k}]=M_k$. Each $M_k$ is integrable because it is square-integrable. Thus this process is a true martingale on the grid, not merely a local martingale.

Worked Example: A Sign-Dependent Coefficient

On the grid $0<1<2$, let $H_0=2$ and $H_1=\mathbf{1}_{\{B_1>0\}}$, where $\mathbf{1}_A$ is the indicator of event $A$. Compute the conditional mean and variance of $M_2$ given $\mathcal{F}_1$, and its unconditional variance. The coefficient is square-integrable because $H_0^2=4$ and $H_1^2\le1$.

Here $M_2=2B_1+\mathbf{1}_{\{B_1>0\}}(B_2-B_1)$. The first term is known given $\mathcal{F}_1$; the increment in the second term is independent of $\mathcal{F}_1$, with mean $0$ and variance $1$. Thus $\mathbb{E}[M_2\mid\mathcal{F}_1]=2B_1$ and $\operatorname{Var}(M_2\mid\mathcal{F}_1)=\mathbf{1}_{\{B_1>0\}}$.

Since $B_1\sim N(0,1)$, symmetry and the zero probability of $B_1=0$ give $\mathbb{P}(B_1>0)=\tfrac12$. Therefore $\mathbb{E}[M_2]=0$, and the conditional second moment is $4B_1^2+\mathbf{1}_{\{B_1>0\}}$. Taking expectations gives $\operatorname{Var}(M_2)=4\cdot1+\tfrac12=\tfrac92$.

Check: the Ito isometry gives $\mathbb{E}\int_0^2H_s^2\,ds=4+\mathbb{E}[\mathbf{1}_{\{B_1>0\}}]=4+\tfrac12=\tfrac92$. This matches the conditional-variance computation.

Worked Example: Future Energy Must Be Averaged

On $0<1<2<3$, set $H_0=1$, $H_1=\mathbf{1}_{\{B_1>0\}}$, and $H_2=B_2$. Each coefficient is square-integrable: their second moments are $1,\tfrac12,$ and $\mathbb{E}[B_2^2]=2$. Find the conditional mean and variance of $M_3-M_1$ given $\mathcal{F}_1$, and the unconditional variance of $M_3$.

The future sum is $M_3-M_1=H_1(B_2-B_1)+B_2(B_3-B_2)$. Each term has conditional mean zero at its own left endpoint; the theorem gives conditional mean zero given $\mathcal{F}_1$. Its conditional variance is $\mathbb{E}[H_1^2+B_2^2\mid\mathcal{F}_1]$. The first term is already known, and $B_2=B_1+(B_2-B_1)$, where the second summand is independent of $\mathcal{F}_1$, has mean $0$, and variance $1$. Thus $\mathbb{E}[B_2^2\mid\mathcal{F}_1]=B_1^2+1$, and $\operatorname{Var}(M_3-M_1\mid\mathcal{F}_1)=\mathbf{1}_{\{B_1>0\}}+B_1^2+1$.

Since $M_1=B_1$ is known given $\mathcal{F}_1$, the conditional variance of $M_3$ is the same. The unconditional mean is zero. Its variance is the variance of the first step plus the expected future conditional variance: $1+\tfrac12+1+1=\tfrac72$. Here the first $1$ is $\mathbb{E}[B_1^2]$, the $\tfrac12$ is $\mathbb{E}[\mathbf{1}_{\{B_1>0\}}]$, and the last $1$ is $\mathbb{E}[B_1^2]$ in the future variance formula.

Check: directly from the Ito isometry, $\mathbb{E}[M_3^2]=\mathbb{E}\int_0^3H_s^2\,ds=1+\tfrac12+2=\tfrac72$, since $\mathbb{E}[B_2^2]=2$. This agrees with the conditional calculation.

Common Pitfall: Treating Future Coefficients as Known

For a single interval, conditioning on its left endpoint makes the coefficient known. But if the conditioning information comes earlier, a later coefficient may still be random. In the sign-dependent example, the variance of the second increment given $\mathcal{F}_1$ is the known indicator $\mathbf{1}_{\{B_1>0\}}$. Given $\mathcal{F}_0$, however, that indicator is not known; its conditional expectation is $\tfrac12$.

A wrong calculation would write the variance contribution from that interval, given $\mathcal{F}_0$, as $\mathbf{1}_{\{B_1>0\}}$. This is not even an $\mathcal{F}_0$-measurable conditional variance. The correction is to average the future energy: $\mathbb{E}[\mathbf{1}_{\{B_1>0\}}\mid\mathcal{F}_0]=\tfrac12$. More generally, the theorem uses $\mathbb{E}[H_i^2h_i\mid\mathcal{F}_{t_k}]$ when the coefficient is not yet known at time $t_k$.

These formulas determine conditional means and variances; they do not, by themselves, assert a distribution for the entire random-integrand integral. Their practical value is that they reduce second-moment calculations to interval energies while preserving exactly which information is available at each time.

Check Your Understanding

In each question, assume Brownian increments are independent of the stated filtration and all step coefficients satisfy the square-integrability conditions above.

If $H$ is $\mathcal{F}_a$-measurable, what are the conditional mean and variance of $H(B_b-B_a)$ given $\mathcal{F}_a$?
For $H_0=1$, $H_1=B_1$ on the grid $0<1<2$, compute the variance of the terminal integral using the Ito isometry.
Why does the conditional variance of a future sum use the conditional expectation of its future energy when some coefficients are not yet known?
In the sign-dependent example, compute $\mathbb{E}[\mathbf{1}_{\{B_1>0\}}\mid\mathcal{F}_0]$.
Explain why zero conditional mean for each step does not make future step coefficients deterministic under earlier conditioning.

Conditional Mean and Variance for Random Step Integrals

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