Derivative of expected value

Question

Let $X\geq 0$ be a real random variable, $Y(t), t\geq 0$ a stochastic process and $A\geq 0$. I want to determine the derivative of $$F(z)=E[e^{X}(A-Y(X))\chi_{\{Y(X)\leq z\}}]$$ $X$ and $Y(X)$ have no density functions and no joint density.I thought about changing integral and derivative to get $$E[e^X(A-Y(X))\frac{d}{dz}\chi_{Y(X)\leq z}]$$. Now I want to calculate the weak derivative of this indicator function so I would get a delta function or a delta measure. Do you have any good idea?

Answer 1

I will assume $A$ is a constant. Let us write

$$ D = D(z) = \frac{\mathrm{d}}{\mathrm{d}z} \mathbb{E}\left[ e^X (A - Y(X)) \mathbf{1}_{\{ Y(X) \leq z \}} \right]. $$

Since it is possible that $D$ does not exist in ordinary sense, we regard $D$ as a weak derivative. To identify the nature of $D$, let $\varphi \in C^{\infty}_c(\mathbb{R})$ be any test function and consider the pairing $\langle \varphi, D \rangle$. Then we get

\begin{align*} \langle \varphi, D \rangle &= - \left< \varphi'(z), \mathbb{E}\left[ e^X (A - Y(X)) \mathbf{1}_{\{ Y(X) \leq z \}} \right] \right> \\ &= - \int_{\mathbb{R}} \varphi'(z) \mathbb{E}\left[ e^X (A - Y(X)) \mathbf{1}_{\{ Y(X) \leq z \}} \right] \, \mathrm{d}z \\ &= \mathbb{E}\left[ e^X (A - Y(X)) \left( - \int_{\mathbb{R}} \varphi'(z) \mathbf{1}_{\{ Y(X) \leq z \}} \, \mathrm{d}z \right) \right] \\ &= \mathbb{E} \left[ e^X (A - Y(X)) \varphi(Y(X)) \right]. \end{align*}

In the third step, we applied Fubini–Tonelli theorem to interchange the order of integration. (Here, we are implicitly assuming that the condition for this theorem is indeed satisfied.)

The above computation confirms that $D$ is indeed the measure defined by

$$ B \quad \mapsto \quad \mathbb{E} \left[ e^X (A - Y(X)) \mathbf{1}_{\{Y(X) \in B\}} \right], $$

which we may write

$$ D = \mathbb{E} \left[ e^X (A - Y(X)) \delta_{Y(X)}(z) \right], $$

at least formally by exploiting the notation. This is essentially what we would expect when we mindlessly interchange the order between integration and differentiation and then invoke the formula $\frac{\mathrm{d}}{\mathrm{d}z} \mathbf{1}_{\{a \leq z\}} = \delta_a(z)$.