Question

Suppose that the measurementXof pressure made bya device in a particular system has the normal distributionwith meanμand variance 1, whereμis the true pressure.Suppose that the true pressureμis unknown but has theuniform distribution on the interval{5,15}. IfX = 8is observed, find the conditional p.d.f. ofμgivenX = 8.

Short answer

The conditional p.d.f of\[\left( {\mu \left| {x = 8} \right.} \right)\]is

\[{f_1}\left( {x\left| \mu \right.} \right) = \frac{{1.0013}}{{\sqrt {2\pi } }}{e^{ - \frac{1}{2}{{\left( {\mu - 8} \right)}^2}}};5 < \mu < 15\]

Step-by-step solution

Given information

X denote the measurement of pressure of a device in a particular system.

X follows N( {μ ,1} and μ denote the true pressure.

μ follows Uniform distribution on [5,15].

Determine the conditional p.d.f of X

The conditional p.d.f of\[\left( {X\left| \mu \right.} \right)\]is

\[{f_1}\left( {x\left| \mu \right.} \right) = \frac{1}{{\sqrt {2\pi } }}{e^{ - \frac{1}{2}{{\left( {x - \mu } \right)}^2}}}\]

Determine the joint p.d.f

Now,\[{f_1}\left( {x\left| \mu \right.} \right){f_1}\left( \mu \right) = \frac{{0.1}}{{\sqrt {2\pi } }}{e^{ - \frac{1}{2}{{\left( {x - \mu } \right)}^2}}};5 \le \mu \le 15\]

Let,

\[\begin{array}{c}{g_1}\left( x \right) = \int\limits_5^{15} {\frac{{0.1}}{{\sqrt {2\pi } }}{e^{ - \frac{1}{2}{{\left( {x - \mu } \right)}^2}}}d\mu } \\ = 0.1\int\limits_5^{15} {\frac{1}{{\sqrt {2\pi } }}{e^{ - \frac{1}{2}{{\left( {x - \mu } \right)}^2}}}d\mu } \\ = 0.1\left[ {\Phi \left( {15 - x} \right) - \Phi \left( {5 - x} \right)} \right]\end{array}\]

If x = 8, then g₁(x) becomes

\[\begin{array}{c}{g_1}\left( x \right) = 0.1\left[ {\Phi \left( {15 - x} \right) - \Phi \left( {5 - x} \right)} \right]\\ = 0.1\left[ {\Phi \left( {15 - 8} \right) - \Phi \left( {5 - 8} \right)} \right]\\ = 0.1\left[ {\Phi \left( 7 \right) - \Phi \left( { - 3} \right)} \right]\\ = 0.1 \times 10.013\\ = 1.0013\end{array}\]

Determine the conditional p.d.f of μ

The conditional p.d.f of\[\left( {\mu \left| {x = 8} \right.} \right)\]is

\[{f_1}\left( {x\left| \mu \right.} \right) = \frac{{1.0013}}{{\sqrt {2\pi } }}{e^{ - \frac{1}{2}{{\left( {\mu - 8} \right)}^2}}};5 < \mu < 15\]