<p>I have to answer this in English as I can not&nbsp;type fast enought&nbsp;in Chinese.</p><p>The 2 state binary process of EDF default simply follows a <em><strong>Bernoulli distribution</strong></em> with parameter <em>p</em> = EDF. Now we know that if <strong><em>X</em></strong> is&nbsp;a&nbsp;Bernoulli distribution which takes value 1 with success probability <span class="texhtml">p, </span><span class="texhtml">E(<em><strong>X</strong></em>) = <em>p*1 + (1-p)*0</em> = <em>p</em>, and Var(<strong><em>X</em></strong>) = E(<em><strong>X</strong></em>^2) - [E(<strong><em>X</em></strong>)]^2 </span><span class="texhtml">= <em>p*(1-p)</em>. </span></p><p><span class="texhtml">Replacing the p in the general model with EDF in your example, you have Variance of EDF = EDF*(1-EDF).</span></p>

楼上正解，

楼上的楼上数学好好啊~~~

thx very much for <font face="Verdana" color="#61b713"><b>qffq</b></font>'s answer!<span style="font-weight: bold;"><br/><br/></span>I made a great mistake for treat EDF as a outcome rather than a probility and fail to recall the Bernoulli Distribution...<span style="font-weight: bold;"><br/><br/>thanks so much for the answer!<br/></span><font face="Verdana" color="#61b713"></font>