Question

Suppose that either of two instruments might be used for making a certain measurement. Instrument 1 yields a measurement whose p.d.f.\({{\bf{h}}_{\bf{1}}}\)is

\({{\bf{h}}_{\bf{1}}}\left( {\bf{x}} \right){\bf{ = }}\left\{ {\begin{align}{}{{\bf{2x}}}&{{\bf{for}}\,{\bf{0 < x < 1}}}\\{\bf{0}}&{{\bf{otherwise}}}\end{align}} \right.\)

Instrument 2 yields a measurement whose p.d.f.\({{\bf{h}}_2}\)is

\({{\bf{h}}_{\bf{2}}}\left( {\bf{x}} \right){\bf{ = }}\left\{ {\begin{align}{}{{\bf{3}}{{\bf{x}}^{\bf{2}}}}&{{\bf{for}}\,{\bf{0 < x < 1}}}\\{\bf{0}}&{{\bf{otherwise}}}\end{align}} \right.\)

Suppose that one of the two instruments is chosen randomly, and a measurement X is made with it.

1. Determine the marginal p.d.f. of X.
2. If the measurement value is\({\bf{X = }}{\raise0.7ex\hbox{\({\bf{1}}\)} \!\mathord{\left/ {\vphantom {{\bf{1}} {\bf{4}}}}\right.\ } \!\lower0.7ex\hbox{\({\bf{4}}\)}}\), what is the probability that instrument 1 was used?

Short answer

1. The marginal p.d.f. of X is, \(\frac{1}{2}{h_1}\left( x \right) + \frac{1}{2}{h_2}\left( x \right)\)
2. The probability of the instrument 1 being used is, 0.8

Step-by-step solution

Given information

The p.d.f. \({h_1}\) is,

\({h_1}\left( x \right) = \left\{ {\begin{align}{}{2x}&{for\,0 < x < 1}\\0&{otherwise}\end{align}} \right.\)

The p.d.f.\({h_2}\)is,

\({h_2}\left( x \right) = \left\{ {\begin{align}{}{3{x^2}}&{for\,0 < x < 1}\\0&{otherwise}\end{align}} \right.\)

Probability calculations

Let Y be the random variable indicating which instrument is being used, i.e.,

\(Y \in \left\{ {1,2} \right\}\)

The pdf of\(X\left| {Y = 1} \right.\)is\({h_1}\left( x \right)\)

The pdf of\(X\left| {Y = 2} \right.\)is\({h_2}\left( x \right)\)

Since an instrument is selected at random i.e.,

\(\Pr \left( {Y = 1} \right) = \frac{1}{2}\)

And,\(\Pr \left( {Y = 2} \right) = \frac{1}{2}\)

It defines the marginal of Y

The joint pdf of X and Y is,

\(f\left( {x,y} \right) = \left\{ {\begin{align}{}{\Pr \left( {Y = 1} \right){h_1}\left( x \right)}&{if}&{y = 1}\\{\Pr \left( {Y = 2} \right){h_2}\left( x \right)}&{if}&{y = 2}\end{align}} \right.\)

(a)

The marginal pdf of X is,

\(\begin{align}\sum\limits_y {f\left( {x,y} \right)} \\ &= \Pr \left( {Y = 1} \right){h_1}\left( x \right) + \Pr \left( {Y = 2} \right){h_2}\left( x \right)\\ &= \frac{1}{2}{h_1}\left( x \right) + \frac{1}{2}{h_2}\left( x \right)\end{align}\)

(b)

The pdf of Y given X is,

\({g_2}\left( {y\left| x \right.} \right) = \frac{{f\left( {x,y} \right)}}{{{f_1}\left( x \right)}}\)

Thus,

\(\begin{align}{g_2}\left( {1\left| x \right.} \right) &= \Pr \left( {Y = 1\left| {X = x} \right.} \right)\\ &= \frac{{f\left( {x,1} \right)}}{{{f_1}\left( x \right)}}\\ &= \frac{{\frac{1}{2}{h_1}\left( x \right)}}{{\frac{1}{2}{h_1}\left( x \right) + \frac{1}{2}{h_2}\left( x \right)}}\end{align}\)

\(\begin{align}{g_2}\left( {2\left| x \right.} \right) &= \Pr \left( {Y = 2\left| {X = x} \right.} \right)\\ &= \frac{{f\left( {x,2} \right)}}{{{f_1}\left( x \right)}}\\ &= \frac{{\frac{1}{2}{h_1}\left( x \right)}}{{\frac{1}{2}{h_1}\left( x \right) + \frac{1}{2}{h_2}\left( x \right)}}\end{align}\)

\(\begin{align}\Pr \left( {Y = 1\left| {X = } \right.\frac{1}{2}} \right) &= {g_2}\left( {1\left| {0.25} \right.} \right)\\ &= \frac{{\frac{1}{2}{h_1}\left( {0.25} \right)}}{{\frac{1}{2}{h_1}\left( {0.25} \right) + \frac{1}{2}{h_2}\left( {0.25} \right)}}\\ &= \frac{{\frac{1}{2} \times 2 \times 0.25}}{{\left( {\frac{1}{2} \times 2 \times 0.25} \right) + \left( {\frac{1}{2} \times 3 \times {{0.25}^2}} \right)}}\\ &= \frac{{0.25}}{{\left( {0.25 + 0.0625} \right)}}\\ &= 0.8\end{align}\)

Therefore, the probability is 0.8