A random variable with normal distribution with unknown mean u and variance 2.25. with a suitable value of u. find the greatest value of P(10.9<X<12.1)
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A random variable with normal distribution with unknown mean u and variance 2.25. with a suitable value of u. find the greatest value of P(10.9<X<12.1)
The value of u which maximises P(10.9<X<12.1) where X~N(u,2.25) isQuote:
Originally Posted by guess
(10.9+12.1)/2=11.5. Now this seems obviouse to me, but let me know if
you need something more.
RonL
ya... u got the ans.. but i does not really know the reason behind it... can you try to explain?
The basic idea is that the normal density is symmetric about the mean, andQuote:
Originally Posted by guess
is decreasing as you move away from the mean, which means that if the
mean is not in the centre of the interval (a,b) that P(a<x<b) can be
increased by moving the mean in the direction of the mid point of the interval.
Alternatively you could write:
$\displaystyle
P(a<x<b)=\int_a^b p(x-u) dx
$
where
$\displaystyle p(x-u)=\frac{1}{\sqrt{2\pi}\ \sigma}e^{-(x-u)^2/2\sigma^2}$
Then:
$\displaystyle
P(a<x<b)=\int_a^u p(x-u) dx+\int_u^b p(x-u) dx
$,
changing the variable of integration in each of the integrals:
$\displaystyle
P(a<x<b)=\int_{a-u}^0 p(x) dx+\int_0^{b-u} p(x) dx
$
The rewriting:
$\displaystyle
P(a<x<b)=-\int_0^{a-u} p(x) dx+\int_0^{b-u} p(x) dx
$
Which can now be differentiated (using the fundamental theorem of calculus)
to give:
$\displaystyle
\frac{d}{du} P(a<x<b)=-p(a-u) + p(b-u)
$
So the stationary points of $\displaystyle P(a<x<b)$ as $\displaystyle u$ varies are the solutions of:
$\displaystyle
p(a-u)=p(b-u)
$.
which are solutions of: $\displaystyle (a-u)^2=(b-u)^2$, which if $\displaystyle a \ne b$ is $\displaystyle u=\frac{a+b}{2}$, and this clearly corresponds to a maximum.
Now I could fill in a few more of the details of the argument, but I won't do
that here. I will just observe that the blindingly obvious can have a relativly
complicated proof :eek:
RonL
