91Ó°ÊÓ

Uniformly distributed over$(a,b)$

$(a,b)$The density function of a Uniform distribution on the interval $(a,b)$is

$f\left(x\right)=\begin{array}{ll}\frac{1}{b−a}& ,x∈(a,b)\\ 0& ,x∉(a,b)\end{array}$

Here we easily see that function f assumes the maximal value of might$>b-a--$lin every point of the interval $(a,b)$so the mode is every point of the interval

Here we easily see that function f assumes the maximal value of might$>b-a-$ lin every point of the interval $(a,b)$so the mode is every point of the interval$(a,b)$

Normal with parameters$\mathrm{μ},\mathrm{σ}$$2$

The Normal distribution with parameters $\mathrm{μ}$and${\mathrm{σ}}^{2}h$as density function

$f\left(x\right)=\frac{1}{\mathrm{σ}\sqrt{2\mathrm{π}}}{e}^{−\frac{(x−\mathrm{μ}{)}^2}{2{\mathrm{σ}}^2}}$

By differentiating, we obtain that

Observe that the expression above is equal to zero if and only if

$X=\mathrm{μ}$

Hence, the only mode is$x=\mathrm{μ}$since it is the only maximum of a function$f$

The only mode i$x=\mathrm{μ}$since it is the only maximum of the function$f$

Exponential with rate λ.

The density function of Exponential distribution with parameter$\mathrm{λ}$is

$f\left(x\right)=\{\begin{array}{ll}\mathrm{λ}{e}^{−\mathrm{λ}x}& ,xâ‰\yen 0\\ 0& ,x<0\end{array}$

Let's show that there are no stationary points on an open interval $(0,∞)$

We have that,

$\frac{df}{dx}=−{\mathrm{λ}}^2{e}^{−\mathrm{λ}x}$

Observe that for all $x∈(0,∞)$we have that $f\text{'}<0$, so there is no stationary point on that interval. Furthermore, because of the fact that the first derivative is negative, $f$it decreases during that interval. On the other hand, f is equal to zero $(-∞,0)$. So, the maximum is assumed in value$0$ Hence, the mode of Exponential distribution is $0$

The mode of Exponential distribution is $0$.