I have a question about an interesting convolution. Suppose the revenues produced by new product development projects are being forecasted. The revenue produced by a new product development project can be considered to be a random variable with an exponential distribution:
$$f_X(x)=\lambda e^{-\lambda x}$$
The forecast of its revenue has an error that is a random variable from a normal distribution:
$$f_Y(y)=\frac{1}{\sqrt{2 \pi (kx)^2}}\exp\left({- \frac{(y-\mu)^2}{2(kx)^2}}\right)$$
where $k$ is a constant. Notice that the standard deviation of the error is proportional to the true revenue, so the size of the error increases with the size of the market. What is the distribution of the forecasts, $Z=X+Y$?
I set up the convolution as:
$$f_Z(z)= \frac{\lambda}{2\pi}\int_{t=0}^\infty \frac{\exp \left( - \lambda t - \frac{(z-t-\mu)^2}{2(kt)^2} \right)}{kt}dt$$
I was unable to do the integration so I solved the integral numerically by using Simpson's Rule. Using reasonable numbers for the parameters and evaluating the integral for various values of Z, I produced the distribution, but one result was strange. I calculated the area under the resulting curve, and rather than being 1, it varied with $k$. As $k$ got larger, the area under the curve got smaller.
Did I set up the convolution incorrectly? Alternatively, should the area under the curve vary with $k$? If so, why does this happen? I appreciate your answers and thoughts. Thank you.