4.1.4 Discussion of numerical results

Since we know that the true value of \(E(Y)=\frac {15}{16}=\allowbreak 0.937\,5\), we see that the

\[ \fbox {95\% confidence interval did contain the true value}\]

We also notice that the relative error in \(\bar {X}\) (the estimate of the expected value) when compared to the true mean \(\mu =\frac {15}{16}\) is calculated as \(\frac {\mu -\bar {X}}{\mu }=\frac {0.937\,5-0.9311632}{0.937\,5}=0.0\allowbreak 06759\,3\simeq 0.7\%\) which is little below the \(1\%\) requirement.

We note that the value of the relative error did not come out exactly \(1\%\) because we used an estimate of the true mean in order to find the sample size needed for the calculation.

The result of the simulation is the estimate of the PDF of \(Y\) which is shown in the plot below. The number of bins used is \(50\). This was determined by trial and error to obtain the most pleasing looking histogram.

We note that the true PDF is given below (derived in the class) and we see from the above plot that the estimated PDF is very close to the analytical PDF.

\[ P\left ( Y=y\right ) =\left \{ \begin {array} [c]{ccc}\frac {3}{4} & & 0\leq y\leq 0.5\\ \frac {1}{4} & & 0.5\leq y\leq 1\\ \frac {1}{2} & & 1<x\leq 2 \end {array} \right . \]