Compute the measurement signal-to-noise ratio that is$\left|\mu \right|/\sigma$, where $\mu =E\left[X\right]$and ${\sigma }^2=Var\left(X\right)$of the following random variables:

$\left(a\right)$Poisson with mean$\lambda$;

$\left(b\right)$binomial with parameters $n$and$p$;

$\left(c\right)$geometric with mean$1/p$;

$\left(d\right)$uniform over$(a,b)$;

$\left(e\right)$exponential with mean$1/\lambda$;

$\left(f\right)$normal with parameters$\mu ,{\sigma }^2$.

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

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

$r=\frac{\left|\mu \right|}{\sigma }$.

Let $X$be a Poisson random variable with a parameter$\lambda$. Then,

$\mu =\lambda$and${\sigma }^2=\lambda$.

Therefore, since $\lambda >0$we have that

$r=\frac{\lambda }{\sqrt{\lambda }}=\sqrt{\lambda }$

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

$\mu =np\text{and}{\sigma }^2=np(1-p)\text{.}$

Since $n\in \mathrm{\mathbb{N}}$and$0\le p\le 1$, we have that $np\ge 0$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,

$\mu =\frac{1}{p},{\sigma }^2=\frac{1-p}{p^2}$

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,

$\mu =\frac{b+a}{2},{\sigma }^2=\frac{(b-a{)}^2}{12}$

and therefore

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

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

$\mu =\frac{1}{\frac{1}{\lambda }}=\lambda ,{\sigma }^2=\frac{1}{{\left(\frac{1}{\lambda }\right)}^2}={\lambda }^2$

and therefore

$r=\frac{\lambda }{\sqrt{{\lambda }^2}}=1$

If we assume that $X$is normally distributed with parameters $\mu$and${\sigma }^2$, then

$r=\frac{\left|\mu \right|}{\sqrt{{\sigma }^2}}=\frac{\left|\mu \right|}{\sigma }$.