$\DeclareMathOperator{\var}{var}\DeclareMathOperator{\cov}{cov}$
The signal-to-noise ratio (SNR) of a random variable quantifies the accuracy of a measurement of a physical quantity. It is defined as $E^2[X]/\var(X)$ and is seen to increase as the mean, which represents the power of the measurement error, that is, $X - E[X]$, decreases. For example, if $X\sim\mathcal{N}(\mu, \sigma^2)$, then $\text{SNR} = \mu^2/\sigma^2$. Determine the SNR if the measurement is $X = A + U$, where $A$ is the true value and $U$ is the measurement error with $U\sim\mathcal{U}(-1/2, 1/2)$. For an SNR of $1000$, what should be $A$?
The mean and variance of a uniform random variable is $E[U] = 0$ and $\var(U) = \frac{1}{12}$. Since the SNR is $1000$, we have that $$ 1000 = \frac{\mu^2}{\sigma^2}\Rightarrow 10\sqrt{10} = \frac{\mu}{\sigma} $$ where $\mu = E[X] = E[A + U] = E[A] + E[U] = E[A]$ and $\sigma^2 = \var(X) = \var(A + U) = \var(A) + \var(U) + 2\cov(A, U) = \var(A) + \frac{1}{12} + \cov(A, U)$.
- Without knowing anything about $A$ how do I find $E[A]$ and $\var(A)$?
- What is meant by what should $A$ be? Is it asking what type of distribution?
In this problem, $A$ is to be treated as a constant. The mean of $A+U$ is just $A$, since $U$ is uniform on $[-1/2,1/2]$, and therefore has mean $0$.
The variance of $A+U$ is just the variance of $U$, which by a standard formula or by a calculation is $\frac{1}{12}$.
For an SNR of $1000$, we want $\frac{A^2}{1/12}=1000$. Solve for $A$.
Remark: To answer your specific questions: 1) The mean of $A$ is $A$ and the variance of $A$ is $0$; 2) The question asks for the value of $A$ that gives SNR equal to $1000$. No distribution is involved.