Question

The definition of the conditional p.d.f. of X given\({\bf{Y = y}}\)is arbitrary if\({{\bf{f}}_{\bf{2}}}\left( {\bf{y}} \right){\bf{ = 0}}\). The reason that this causes no serious problem is that it is highly unlikely that we will observe Y close to a value\({{\bf{y}}_{\bf{0}}}\)such that\({{\bf{f}}_{\bf{2}}}\left( {{{\bf{y}}_{\bf{0}}}} \right){\bf{ = 0}}\). To be more precise, let\({{\bf{f}}_{\bf{2}}}\left( {{{\bf{y}}_{\bf{0}}}} \right){\bf{ = 0}}\), and let\({{\bf{A}}_{\bf{0}}}{\bf{ = }}\left( {{{\bf{y}}_{\bf{0}}}{\bf{ - }} \in {\bf{,}}{{\bf{y}}_{\bf{0}}}{\bf{ + }} \in } \right)\). Also, let\({{\bf{y}}_{\bf{1}}}\)be such that\({{\bf{f}}_{\bf{2}}}\left( {{{\bf{y}}_{\bf{1}}}} \right){\bf{ > 0}}\), and let\({{\bf{A}}_{\bf{1}}}{\bf{ = }}\left( {{{\bf{y}}_{\bf{1}}}{\bf{ - }} \in {\bf{,}}{{\bf{y}}_{\bf{1}}}{\bf{ + }} \in } \right)\). Assume that\({{\bf{f}}_{\bf{2}}}\)is continuous at both\({{\bf{y}}_{\bf{0}}}\)and\({{\bf{y}}_{\bf{1}}}\).

Show that

\(\mathop {{\bf{lim}}}\limits_{ \in \to {\bf{0}}} \,\frac{{{\bf{Pr}}\left( {{\bf{Y}} \in {{\bf{A}}_{\bf{0}}}} \right)}}{{{\bf{Pr}}\left( {{\bf{Y}} \in {{\bf{A}}_{\bf{1}}}} \right)}}{\bf{ = 0}}{\bf{.}}\)

That is, the probability that Y is close to\({{\bf{y}}_{\bf{0}}}\)is much smaller than the probability that Y is close to\({{\bf{y}}_{\bf{1}}}\).

Short answer

\(\mathop {\lim }\limits_{ \in \to 0} \frac{{\Pr \left( {Y \in {A_i}} \right)}}{{\Pr \left( {Y \in {A_1}} \right)}} = 0\)

Step-by-step solution

Given information

The conditional p.d.f. of X given\(Y = y\)is arbitrary if\({f_2}\left( y \right) = 0\)

Y is close to\({y_0}\)such that\({f_2}\left( {{y_0}} \right) = 0\).

And let,

\({A_0} = \left( {{y_0} - \in ,{y_0} + \in } \right)\)

And\({y_1}\)is such that,\({f_2}\left( {{y_1}} \right) > 0\), and\({A_1} = \left( {{y_1} - \in ,{y_1} + \in } \right)\)

\({f_2}\) is continuous at both \({y_0}\) and \({y_1}\)

Necessary calculation

Let, \({F_2}\) be the c.d.f. of Y.

Since\({f_2}\)is continuous at both\({y_0}\,and\,{y_1}\), we can write, for\(i = 0,1\)

\(\begin{align}\Pr \left( {Y \in {A_i}} \right) &= {F_2}\left( {{y_i} - \in } \right) - {F_2}\left( {{y_i} - \in } \right)\\ &= 2 \in {f_2}\left( {{{\left( {{y_i}} \right)}^\prime }} \right),\end{align}\)

Where,\({y_i}^\prime \)is within\( \in \)of\({y_i}\).

This last equation follows from the mean value theorem of calculus.

So,

\(\frac{{\Pr \left( {Y \in {A_i}} \right)}}{{\Pr \left( {Y \in {A_1}} \right)}} = \frac{{{f_2}\left( {{{\left( {{y_0}} \right)}^\prime }} \right)}}{{{f_2}\left( {{{\left( {{y_1}} \right)}^\prime }} \right)}}\)

Since\({f_2}\)is continuous,

\(\mathop {\lim }\limits_{ \in \to 0} {f_2}\left( {{{\left( {{y_i}} \right)}^\prime }} \right) = {f_2}\left( {{y_i}} \right)\)

So,

\(\begin{align}\mathop {\lim }\limits_{ \in \to 0} \frac{{\Pr \left( {Y \in {A_i}} \right)}}{{\Pr \left( {Y \in {A_1}} \right)}} &= \mathop {\lim }\limits_{ \in \to 0} \frac{{{f_2}\left( {{{\left( {{y_0}} \right)}^\prime }} \right)}}{{{f_2}\left( {{{\left( {{y_1}} \right)}^\prime }} \right)}}\\ &= \frac{{\mathop {\lim }\limits_{ \in \to 0} {f_2}\left( {{{\left( {{y_0}} \right)}^\prime }} \right)}}{{\mathop {\lim }\limits_{ \in \to 0} {f_2}\left( {{{\left( {{y_1}} \right)}^\prime }} \right)}}\\ &= \frac{{{f_2}\left( {{y_0}} \right)}}{{{f_2}\left( {{y_1}} \right)}}\\ &= \frac{0}{{{f_2}\left( {{y_1}} \right)}}\\ &= 0\end{align}\)

Therefore, \(\mathop {\lim }\limits_{ \in \to 0} \frac{{\Pr \left( {Y \in {A_i}} \right)}}{{\Pr \left( {Y \in {A_1}} \right)}} = 0\)