No, but I can tell you why it feels like you are double counting.
Consider a cash flow $\tilde{x}=\tilde{x}(t,\mu,\sigma^2)$ to be received in the future. While many cash flows lack a first moment and so no defined mean or variance, let us assume at least the second moment is defined to make the discussion simple. Implicitly, your assumption of an expectation would require a first moment to exist. Indeed to make this easier, let us assume normality.
If $t$ is time; $\mu$ the center of location; and, $\sigma^2$ the scale parameter, then we can talk about a rate. From the formula $$\rho=\frac{\mathbb{E}(\tilde{x}(\mu(t),\sigma^2(t)))}{1+d(\mu(t),\sigma^2(t))}.$$ Your assumption of additivity is problematic, while it is often used as an approximation, if you think about it for a second you will see why. Instead, I am defining $$d=(1+r(\mu_0,\sigma^2_0))(1+\tau(t))$$ because I am using $t$ for time instead of risk. $\tau(t)$ is the function that maps the premium as a function of time to discount a certain cash flow.
We will adopt a Frequentist interpretation of probability to make this simple. Using multiplication all the constants are on the left and the expectation of the random variables are on in the center when we rearrange it as $$\rho(1+d(\mu(t),\sigma^2(t))=\mathbb{E}(\tilde{x}(\mu(t),\sigma^2(t)))=\rho\mu(t).$$
The present value cannot be stochastic as it is known by observation. Since $\mu(t)$ and $\sigma^2(t)$ are constants by definition in the Frequentist interpretation of probability nothing on the left or right contains any randomness at all. Only the center is random. That randomness is averaged out over the sample space so that only the point is left.
You could logically take this one step further and argue that $\mu(t)=\mu(t,\sigma^2(t))$. It feels like it is double counted because the mean is a function of the variance and the cash flow is a function of the mean and the variance. The rate is a function of the risk-adjusted mean, so it is a function of the mean and the variance.
You could dissolve the mean and convert it into a pure function of variance and time, and then only the scale parameter would exist in the numerator and the denominator.
$$\rho=\frac{\mathbb{E}(\tilde{x}(\sigma^2(t),t))}{1+d(\sigma^2(t),t)}$$
You are also missing the observation that $\mu(t,\sigma^2(t))$ is similar to an expenditure function and not merely a center of location.
If you step back one more unit of time, to before time zero, then $\rho$ becomes stochastic as well because $\tilde{\rho}=\rho$ if and only if $\mathbb{E}[\mathcal{U}(\tilde{x})]>\mathcal{U}(\tilde{\rho}=0)$, where $\mathcal{U}()$ is a utility function.