Difference between revisions of "Solution3 Q1 CS1 2013QE" - Rhea

@@ Line 1: / Line 1: @@
-[[Category:ECE PhD QE CNSIP 2013 Problem1.1]][[Category:ECE PhD QE CNSIP 2013 Problem1.1]][[Category:ECE PhD QE CNSIP 2013 Problem1.1]][[Category:ECE PhD QE CNSIP 2013 Problem1.1]][[Category:ECE PhD QE CNSIP 2013 Problem1.1]][[Category:ECE PhD QE CNSIP 2013 Problem1.1]]
-=Solution3=
-<p>
-First, we define a Bernoulli random variable
-</p>
-<p><span class="mwe-math-fallback-source-inline tex" dir="ltr">$ X = \left\{ \begin{array}{ll}  0, &amp; the&nbsp;change&nbsp;over&nbsp;does&nbsp;not&nbsp;occur\\  1, &amp; the&nbsp;change&nbsp;over&nbsp;occurs  \end{array} \right. $</span>
-</p><p>Then we can compute
-</p><p><span class="mwe-math-fallback-source-inline tex" dir="ltr">$P(X = 1) = P(1-P)+(1-P)P = P-{P}^{2}+P-{P}^2=2P-2{P}^{2} $</span>
-</p><p><span class="mwe-math-fallback-source-inline tex" dir="ltr">$P(X = 0) = P•P+(1-P)(1-P) = {P}^{2}+1-2P+{P}^2 $</span>
-</p><p>Define Y as the number of changes occurred in n flips, there exists at most n-1 changes
-</p><p><span class="mwe-math-fallback-source-inline tex" dir="ltr">$E(Y)=E(\sum_{i=1}^{n-1}X_i)=\sum_{i=1}^{n-1}E(X_i) $</span>
-</p><p><span class="mwe-math-fallback-source-inline tex" dir="ltr">$E(X_i)=p(X_i=1)=p(1-p)+(1-p)p=2p(1-p) $</span>
-</p><p><span class="mwe-math-fallback-source-inline tex" dir="ltr">$ E(Y)=2(n-1)p(1-p) $</span>.
-</p>
-</p>[[ ECE PhD QE CNSIP 2013 Problem1.1|Back to ECE PhD QE CNSIP 2013 Problem1.1]]

Latest revision as of 22:39, 22 February 2017