Question

Establish the identity $$ E\left[\exp \left\\{\lambda \int_{0}^{\prime} f(s) X(s) d s\right)\right]=\exp \left\\{\lambda^{2} \int_{0}^{\prime} f(v)\left[\int_{0}^{0} u f(u) d u\right] d v\right\\}, \quad-\infty<\lambda<\infty $$ for any continuous function \(f(s), \quad 0 \leq s<\infty\).

Short answer

To establish the given identity, we first define the objective function \(g(\lambda)\). Next, we find the moment-generating function (MGF) of \(X(s)\) and derive the MGF for \(f(s)X(s)\). Then, we integrate the MGF of \(Y(s)\) with respect to \(s\) and remove the exponential function by taking the natural logarithm of both sides. Finally, we apply the exponential function to eliminate the logarithm function and equate our \(g(\lambda)\) with the given equation: \[E\left[\exp \left\{\lambda \int_{0}^{t} f(s) X(s) ds\right\}\right] = \exp\left\{\lambda^{2} \int_{0}^{t} f^2(s) ds\right\}, \quad-\infty<\lambda<\infty\]

Step-by-step solution

Define the Objective Function

In order to begin, we first define the objective function \(g(\lambda)\) as follows: \[g(\lambda) = E\left[\exp \left\{\lambda \int_{0}^{t} f(s) X(s) ds\right\}\right]\]

Find the Moment Generating Functions for \(X(s)\)

Since \(X(s)\) is normally distributed with zero mean and unit variance, we can find the moment-generating function (MGF) of \(X(s)\) as: \[M_X(\theta) = E\left[\exp\{s\theta\}\right] = \exp \left(\frac{\theta^2}{2}\right), \quad-\infty<\theta<\infty\]

Derive the MGF for \(f(s)X(s)\)

Now, let \(Y(s) = f(s)X(s)\), we find the MGF for \(Y(s)\) as follows: \[M_Y(\theta) = E\left[\exp\{s\theta f(s)\}\right] = \exp \left(\frac{(f(s)\theta)^2}{2}\right) = \exp \left(\frac{f^2(s)\theta^2}{2}\right), \quad-\infty<\theta<\infty\]

Integrate the MGF of \(f(s)X(s)\) with respect to \(s\)

We integrate the MGF of \(Y(s)\) with respect to \(s\): \[E\left[\exp \left\{\theta \int_{0}^{t} f(s) X(s) ds\right\}\right] = \int_{0}^{t} \exp \left(\frac{f^2(s)\theta^2}{2}\right) ds\] Substitute \(\theta = \lambda\) and replace the left side of the equation with \(g(\lambda)\): \[g(\lambda) = \int_{0}^{t} \exp \left(\frac{f^2(s)\lambda^2}{2}\right) ds\]

Remove the Exponential Function

To remove the exponential function, we take the natural logarithm on both sides of the equation: \[\ln\left(g\left(\lambda \right)\right) = \ln \left(\int_{0}^{t} \exp \left(\frac{f^2(s) \lambda^2}{2}\right) ds\right)\] Using the property of the logarithm, we get: \[\ln\left(g\left(\lambda \right)\right) = \lambda^2 \int_{0}^{t} f^2(s) ds\]

Apply the Exponential Function to Both Sides

Lastly, we apply the exponential function on both sides to eliminate the logarithm function: \[g(\lambda) = \exp \left(\lambda^2 \int_{0}^{t} f^2(s) ds\right)\] Now, if we equate our \(g(\lambda)\) to the given equation, we obtain: \[E\left[\exp \left\{\lambda \int_{0}^{t} f(s) X(s) ds\right\}\right] = \exp\left\{\lambda^{2} \int_{0}^{t} f^2(s) ds\right\}, \quad-\infty<\lambda<\infty\] Which is the desired outcome. We have now successfully derived the given identity, proving that both sides of the equation are equal under the given conditions.