91Ó°ÊÓ

The measurement signal-to-noise ratio is$\left|\mathrm{μ}\right|/\mathrm{σ}$, where $\mathrm{μ}=E\left[X\right]$and${\mathrm{σ}}^2=Var\left(X\right)$.

Assume that the random variable $X$has mean $\mathrm{μ}=E\left[X\right]$and variance${\mathrm{σ}}^2=Var\left(X\right)$. The measurement signal-to-noise ratio $X$is defined as

$r=\frac{\left|\mathrm{μ}\right|}{\mathrm{σ}}$.

Let $X$be a Poisson random variable with a parameter$\mathrm{λ}$. Then,

$\mathrm{μ}=\mathrm{λ}$and${\mathrm{σ}}^2=\mathrm{λ}$.

Therefore, since $\mathrm{λ}>0$we have that

$r=\frac{\mathrm{λ}}{\sqrt{\mathrm{λ}}}=\sqrt{\mathrm{λ}}$

Let $X$be a binomial random variable with parameters$(n,p)$. Then,

$\mathrm{μ}=np\text{and}{\mathrm{σ}}^2=np(1-p)\text{.}$

Since $n∈\mathrm{â„•}$and$0≤p≤1$, we have that $npâ‰\yen 0$and therefore

$r=\frac{np}{\sqrt{np(1-p)}}=\sqrt{\frac{np}{(1-p)}}$

Assume that$0<p<1$. Let $X$be a geometric random variable with parameterslocalid="1649910508197" role="math" $1/p$. Then,

$\mathrm{μ}=\frac{1}{p},{\mathrm{σ}}^2=\frac{1-p}{p^2}$

and therefore

$r=\frac{\frac{1}{p}}{\sqrt{\frac{1-p}{p^2}}}=\frac{1}{\sqrt{1-p}}$

Let $X$be uniformly distributed over$(a,b)$. Then,

$\mathrm{μ}=\frac{b+a}{2},{\mathrm{σ}}^2=\frac{(b-a{)}^2}{12}$

and therefore

$r=\frac{\frac{|b+a|}{2}}{\sqrt{\frac{(b-a{)}^2}{12}}}=\frac{\sqrt{3}|b+a|}{|b-a|}.$

Assume that $\mathrm{λ}>0$and let $X$be an exponential random variable with a parameter$1/\mathrm{λ}$. Then,

$\mathrm{μ}=\frac{1}{\frac{1}{\mathrm{λ}}}=\mathrm{λ},{\mathrm{σ}}^2=\frac{1}{{\left(\frac{1}{\mathrm{λ}}\right)}^2}={\mathrm{λ}}^2$

and therefore

$r=\frac{\mathrm{λ}}{\sqrt{{\mathrm{λ}}^2}}=1$

If we assume that $X$is normally distributed with parameters $\mathrm{μ}$and${\mathrm{σ}}^2$, then

$r=\frac{\left|\mathrm{μ}\right|}{\sqrt{{\mathrm{σ}}^2}}=\frac{\left|\mathrm{μ}\right|}{\mathrm{σ}}$.