5
$\begingroup$

Given the following definitions:

$X \sim \mathcal{N}(\bar{x}, \sigma_{0}^{2})$, and $W_i \sim \mathcal{N}(0, \sigma^{2})$, $i \in \{1,2\}$ and $E[W_1W_2]=\rho \sigma^2$. $X$ and $W_i$ are independent.

Two measurements $Z_1$ and $Z_2$ are performed $$Z_i = X + W_i$$ What is the distribution of vector RV $Z = [Z_1 \ Z_2]^{T}$?

I tried to get the distribution by calculating CDF of $Z$

$$F_Z(z_1,z_2)=P\{Z_1 \leq z_1,Z_2 \leq z_2\} = P\{Z_1 \leq z_1 | Z_2 \leq z_2\} P\{Z_2 \leq z_2\}$$ I know how to calculate $P\{Z_2 \leq z_2\}$ $$Z_2 \sim \mathcal{N}(\bar{x}, \sigma_{0}^{2} + \sigma^{2})$$ I don't know how to calculate conditional probability $P\{Z_1 \leq z_1 | Z_2 \leq z_2\}$.

Is there different approach to find out how $Z$ is distributed?

$\endgroup$
  • $\begingroup$ $\bar{x}$ would normally indicate a sample mean rather than a population mean (for which $\mu$ would most typically be used). It seems odd to avoid the Greek letter for the mean but to use it for the variance. $\endgroup$ – Glen_b Jul 26 at 23:21
  • $\begingroup$ I follow the convention in the book I currently study. Yaakov & Li, X.‐Rong & Kirubarajan, Thia. (2004). Estimation with Applications to Tracking and Navigation: Theory, Algorithms and Software. My question is related to exercise from the book in which x bar was used for mean and sigma for variance. $\endgroup$ – Valjean Jul 27 at 6:24
  • $\begingroup$ That the book (inconsistently!) flouts convention bodes very poorly for the book (for example it hints at substantial deficits of understanding on the part of the authors, though perhaps it points to a lack of clarity of exposition instead); it doesn't make it seem any less odd. $\endgroup$ – Glen_b Jul 28 at 12:44
  • $\begingroup$ For my question you are right, I should have used Greek letter for the expectation. In the context of state estimation. Where $x$ represents state or random process it is common to use $\bar{x}$. $\endgroup$ – Valjean Jul 28 at 19:31
  • $\begingroup$ It might be worth noting the choice of notation is from the book; I wouldn't change it. $\endgroup$ – Glen_b Jul 29 at 1:18
4
$\begingroup$

It seems you assume $X$ is independent of $W_i$, and $W_i$ are jointly normal. Following this fact, $Z_i$ become jointly normal, and you can use the conditional distribution formula to find the joint PDF of $Z_i$. A similar approach for finding the joint PDF would be directly calculating the mean and covariance vector for $[Z_1, Z_2]$ ($Z$ is jointly normal because it's a linear transform of the random vector $[X,W_1,W_2]$). However, multivariate normal CDF has no closed form.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I don't want closed form solution. It is fine to show that $Z$ is Gaussian and hence cdf is known. Or express the solution with cdf of Gaussian distributions. How it follows from the fact that $Z_1$ and $Z_2$ are Gaussian that $Z$ is Gaussian? $\endgroup$ – Valjean Jul 26 at 12:36
  • $\begingroup$ If $W_i$ are jointly normal (so as $X$ since it's independent), by the properties of jointly normal RVs, a linear transformation on the vector $[X,W_1,W_2]$, which is $[Z_1,Z_2]$ in this case, is also a jointly normal vector. $\endgroup$ – gunes Jul 26 at 13:04
  • $\begingroup$ Now I understand. I feel linear transformation argument should be added to your answer. $[X,W_1,W_2]=Y \sim \mathcal{N}(\mathbf{m}=[\bar{x} \ 0 \ 0]',P=\begin{bmatrix} \sigma_{0}^{2} & 0 & 0 \\ 0 & \sigma^2 & \rho \sigma^2 \\ 0 & \rho\sigma^2 & \sigma^2 \end{bmatrix})$ and $Z=BY$ where $B=\begin{bmatrix} 1 & 1 & 0 \\ 1 &0 &1\end{bmatrix}$ $\implies$ $Z \sim \mathcal{N}(B\mathbf{m},BPB')$ $\endgroup$ – Valjean Jul 26 at 16:19
  • $\begingroup$ Correct formulation! I'll mention it more explicitly. $\endgroup$ – gunes Jul 26 at 16:21
  • $\begingroup$ I corrected the formulation. $\endgroup$ – Valjean Jul 26 at 16:30
1
$\begingroup$

Addition to the previous answer. You can approximate a bivariate normal distribution with an arbitrary mean vector ${\bf \mu}$ and covariance matrix $\Sigma$ by the eigen-vector transformation of the covariance matrix (see Appendix A.2 in (n-dimensional normal distribution). This yields an n-dimensional normal distribution with ${\bf \mu}\prime = (0,\ldots,0)^T$ and the covariance matrix $\Sigma \prime =I$ $\; \; -$ the identity matrix.

Using a numerical approximation formula like the bivariate normal integral, you can compute probabilities in the bivariate normal distribution.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.