# uncertainty propagation - a distribution function that is conditional to a probability distribution

The title may be weird, but I can explain it more detailedly.

First, there is a variable $\hat{x}$ that follows Gaussian distribution $N(\psi, \sigma_1)$. When given $\hat{x}$, there is another variable y that also follows the Gaussian distribution whose mean is at $\hat{x}$ and the standard deviation is $\sigma_2$, $N(\hat{x}, \sigma_2)$. Now I would like to find the overall standard deviation of the y, and even its probability distribution function. I can imagine that the overall y should have a mean at $\psi$ too, and its standard deviation (or variance) should be somehow related to $\sigma_1 + \sigma_2$.

One practical problem to help the understanding. For a machine system which has sensing and execution modules. We want to move the machine to a target position, whose actual location is $\psi$. The sensing module has some uncertainty following the Gaussian distribution $N(\psi, \sigma_1)$, and senses the target location at $\hat{x}$. And this location $\hat{x}$ is given to the execution module, however, the execution module has uncertainty too, which will be the $N(\hat{x}, \sigma_2)$. Thus, at the end, the machine will end up at some location y. Now, I want to know the overall variance of y and even its probability distribution function.

I tried to compute it through the conditional probability as $P(y) = \sum\nolimits_{\hat{x}} P(y|\hat{x}）P(\hat{x})$, and $P(x)$ is acturely the accurate probabality from $x-\epsilon$ to $x+\epsilon$, $\epsilon$ is a vary small number that close to zero. But then I lost in the math.

• Consider $Y$ and $Z$ independent random variables with normal distributions where $Y$ had mean $0$ and standard deviation $\sigma_1$ and $Z$ had mean $0$ and standard deviation $\sigma_2$. The I think your example has the same distribution as $\psi+Y+Z$ Jun 18 '17 at 22:29
Your approach is ok, but you need to integrate. We have $$f_Y(y) = \int f_{Y|X}(y|x)f_X(x) dx = \int_{-\infty}^\infty \frac{e^{-\frac{1}{2\sigma_1^2}(x-\psi)^2}}{\sqrt{2\pi\sigma_1^2}} \frac{e^{-\frac{1}{2\sigma_2^2}(y-x)^2}}{\sqrt{2\pi\sigma_2^2}}dx$$ so it's a matter of doing this integral, which can be done by completing the square.
But as Henry said in the comments, there is no reason other than practice to go through all of the work. Since we can get a mean $\mu$ normal by adding $\mu$ to a mean zero normal, the distribution of $Y$ will have to be the distribution of $\hat X+Y'$ where $Y'$ is a normal with mean zero and standard deviation $\sigma_2.$ (Note that this result doesn't require $\hat X$ to be normal, or any assumptions about it at all.) So by addition of normal variables, $Y$ is distributed as $N(\psi, \sigma_1^2+\sigma_2^2).$