exercise:A917bb4aad: Difference between revisions
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(Created page with "<div class="d-none"><math> \newcommand{\NA}{{\rm NA}} \newcommand{\mat}[1]{{\bf#1}} \newcommand{\exref}[1]{\ref{##1}} \newcommand{\secstoprocess}{\all} \newcommand{\NA}{{\rm NA}} \newcommand{\mathds}{\mathbb}</math></div> Let <math>X</math> and <math>Y</math> be random variables with values in <math>\{1,2,3,4,5,6\}</math> with distribution functions <math>p_X</math> and <math>p_Y</math> given by <math display="block"> \begin{eqnarray*} p_X(j) &=& a_j\ , \\ p_Y(j)...") |
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Let <math>X</math> and <math>Y</math> be random variables with values in | |||
<math>\{1,2,3,4,5,6\}</math> with distribution functions <math>p_X</math> and <math>p_Y</math> given by | <math>\{1,2,3,4,5,6\}</math> with distribution functions <math>p_X</math> and <math>p_Y</math> given by | ||
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<ul><li> Find the ordinary generating functions <math>h_X(z)</math> and <math>h_Y(z)</math> for these | <ul style="list-style-type:lower-alpha"><li> Find the ordinary generating functions <math>h_X(z)</math> and <math>h_Y(z)</math> for these | ||
distributions. | distributions. | ||
</li> | </li> | ||
Latest revision as of 00:45, 15 June 2024
Let [math]X[/math] and [math]Y[/math] be random variables with values in [math]\{1,2,3,4,5,6\}[/math] with distribution functions [math]p_X[/math] and [math]p_Y[/math] given by
[[math]]
\begin{eqnarray*}
p_X(j) &=& a_j\ , \\
p_Y(j) &=& b_j\ .
\end{eqnarray*}
[[/math]]
- Find the ordinary generating functions [math]h_X(z)[/math] and [math]h_Y(z)[/math] for these distributions.
- Find the ordinary generating function [math]h_Z(z)[/math] for the distribution [math]Z = X + Y[/math].
- Show that [math]h_Z(z)[/math] cannot ever have the form
[[math]] h_Z(z) = \frac{z^2 + z^3 +\cdots+ z^{12}}{11}\ . [[/math]]
Hint: [math]h_X[/math] and [math]h_Y[/math] must have at least one nonzero root, but [math]h_Z(z)[/math] in the form given has no nonzero real roots. It follows from this observation that there is no way to load two dice so that the probability that a given sum will turn up when they are tossed is the same for all sums (i.e., that all outcomes are equally likely).