<div dir="ltr">faca o estudo de funcao... sendo super "simplista", pergunte-se:<div><br></div><div>qual "lambda" faz 1/lambda = 0?</div><div><br></div><div>vc deveria responder "infinito"...</div>

<div><br></div><div>agora num cenario de amostra finita: entre 5 numeros, x1...x5, qual e' o numero "mais prox do infinito"? e vc deveria respnder max(x)...</div></div><div class="gmail_extra"><br><br><div class="gmail_quote">

Em 29 de junho de 2014 09:22,  <span dir="ltr"><<a href="mailto:andrebvs@bol.com.br" target="_blank">andrebvs@bol.com.br</a>></span> escreveu:<br><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">

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<div class="gmail_default" style="font-family:'trebuchet ms',sans-serif">Não entendi a parte da função de verossimilhança, tentei desenvolver mas não chego no resultado.<br><br>L(<span style="color:#545454;font-family:arial,sans-serif;font-size:small;line-height:18.200000762939453px">λ;x</span>) = <span style="color:#545454;font-family:arial,sans-serif;font-size:small;line-height:18.200000762939453px">Π f(xi) = (1/</span><span style="color:#545454;font-family:arial,sans-serif;font-size:small;line-height:18.200000762939453px">λ) . </span><span style="font-size:small;color:#545454;font-family:arial,sans-serif;line-height:18.200000762939453px">(1/</span><span style="font-size:small;color:#545454;font-family:arial,sans-serif;line-height:18.200000762939453px">λ) . </span><span style="font-size:small;color:#545454;font-family:arial,sans-serif;line-height:18.200000762939453px">(1/</span><span style="font-size:small;color:#545454;font-family:arial,sans-serif;line-height:18.200000762939453px">λ) . </span><span style="font-size:small;color:#545454;font-family:arial,sans-serif;line-height:18.200000762939453px">(1/</span><span style="font-size:small;color:#545454;font-family:arial,sans-serif;line-height:18.200000762939453px">λ) . </span><span style="font-size:small;color:#545454;font-family:arial,sans-serif;line-height:18.200000762939453px">(1/</span><span style="font-size:small;color:#545454;font-family:arial,sans-serif;line-height:18.200000762939453px">λ) = </span><span style="font-size:small;color:#545454;font-family:arial,sans-serif;line-height:18.200000762939453px">(1/</span><span style="font-size:small;color:#545454;font-family:arial,sans-serif;line-height:18.200000762939453px">λ)^5<br>

<br>Aplicando o logarítmo na função </span>L(<span style="font-size:small;color:#545454;font-family:arial,sans-serif;line-height:18.200000762939453px">λ;x</span>)<span style="font-size:small;color:#545454;font-family:arial,sans-serif;line-height:18.200000762939453px">, teremos:<br>

<br>ln[</span>L(<span style="font-size:small;color:#545454;font-family:arial,sans-serif;line-height:18.200000762939453px">λ;x</span>)<span style="color:#545454;font-family:arial,sans-serif;font-size:small;line-height:18.200000762939453px">] = ln[</span><span style="font-size:small;color:#545454;font-family:arial,sans-serif;line-height:18.200000762939453px">(1/</span><span style="font-size:small;color:#545454;font-family:arial,sans-serif;line-height:18.200000762939453px">λ)^5</span><span style="color:#545454;font-family:arial,sans-serif;font-size:small;line-height:18.200000762939453px">] = 5.[ln(1) - ln</span><span style="color:#545454;font-family:arial,sans-serif;font-size:small;line-height:18.200000762939453px">(</span><span style="color:#545454;font-family:arial,sans-serif;font-size:small;line-height:18.200000762939453px">λ</span><span style="color:#545454;font-family:arial,sans-serif;font-size:small;line-height:18.200000762939453px">)] = </span><span style="font-size:small;color:#545454;font-family:arial,sans-serif;line-height:18.200000762939453px">5.[0 - ln</span><span style="font-size:small;color:#545454;font-family:arial,sans-serif;line-height:18.200000762939453px">(</span><span style="font-size:small;color:#545454;font-family:arial,sans-serif;line-height:18.200000762939453px">λ</span><span style="font-size:small;color:#545454;font-family:arial,sans-serif;line-height:18.200000762939453px">)] = </span><span style="font-size:small;color:#545454;font-family:arial,sans-serif;line-height:18.200000762939453px">- 5ln</span><span style="font-size:small;color:#545454;font-family:arial,sans-serif;line-height:18.200000762939453px">(</span><span style="font-size:small;color:#545454;font-family:arial,sans-serif;line-height:18.200000762939453px">λ</span><span style="font-size:small;color:#545454;font-family:arial,sans-serif;line-height:18.200000762939453px">)<br>

<br>Derivando </span><span style="font-size:small;color:#545454;font-family:arial,sans-serif;line-height:18.200000762939453px">ln[</span>L(<span style="font-size:small;color:#545454;font-family:arial,sans-serif;line-height:18.200000762939453px">λ;x</span>)<span style="font-size:small;color:#545454;font-family:arial,sans-serif;line-height:18.200000762939453px">] </span><span style="color:#545454;font-family:arial,sans-serif;font-size:small;line-height:18.200000762939453px">e igualando a zero, teremos:<br>

<br><span style="text-decoration:underline">d</span></span><span style="text-decoration:underline"><span style="font-size:small;color:#545454;font-family:arial,sans-serif;line-height:18.200000762939453px;text-decoration:underline">ln[</span>L(<span style="font-size:small;color:#545454;font-family:arial,sans-serif;line-height:18.200000762939453px;text-decoration:underline">λ;x</span>)</span><span style="font-size:small;color:#545454;font-family:arial,sans-serif;line-height:18.200000762939453px"><span style="text-decoration:underline">]</span> = -5(</span><span style="font-size:small;color:#545454;font-family:arial,sans-serif;line-height:18.200000762939453px">1/</span><span style="font-size:small;color:#545454;font-family:arial,sans-serif;line-height:18.200000762939453px">λ</span><span style="font-size:small;color:#545454;font-family:arial,sans-serif;line-height:18.200000762939453px">) = 0   ???????<br>

     d</span><span style="color:#545454;font-family:arial,sans-serif;font-size:small;line-height:18.200000762939453px">λ</span><span style="font-size:small;color:#545454;font-family:arial,sans-serif;line-height:18.200000762939453px"><br>

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<div><br><strong>De:</strong> <a href="mailto:walmeszeviani@gmail.com" target="_blank">walmeszeviani@gmail.com</a><br><strong>Enviada:</strong> Domingo, 29 de Junho de 2014 01:29<br><strong>Para:</strong> <a href="mailto:r-br@listas.c3sl.ufpr.br" target="_blank">r-br@listas.c3sl.ufpr.br</a><br>

<strong>Assunto:</strong> [R-br] Questão de concurso verossimilhança...<br><br></div><div><div class="h5">
<div dir="ltr">
<div class="gmail_default" style="font-family:trebuchet ms,sans-serif">Para determinar a variância você pode partir do fato que<br><br>V(X) = E(X^2) - (E(X))^2.<br><br></div>
<div class="gmail_default" style="font-family:trebuchet ms,sans-serif">Tem-se que<br><br></div>
<div class="gmail_default" style="font-family:trebuchet ms,sans-serif">E(X) = \int x f(x) dx = \int_{0}^{\lambda} x/\lambda dx = \lambda/2<br><br></div>
<div class="gmail_default" style="font-family:trebuchet ms,sans-serif">Tem-se que E(X^2), pelo mesmo caminho é \lambda^2/3.<br><br></div>
<div class="gmail_default" style="font-family:trebuchet ms,sans-serif">Com isso V(X) = \lambda^2/3 - \lambda^2/4.<br><br></div>
<div class="gmail_default" style="font-family:trebuchet ms,sans-serif">A função de verossimilhança é<br><br>\prod_{1 até n} 1/lambda para x<\lambda, 0 se pelo menos um x>lambda.<br><br></div>
<div class="gmail_default" style="font-family:trebuchet ms,sans-serif">Do estudo da função tem-se que o seu máximo ocorre, considerando a amostra, quando lambda=24 (o maior valor).<br> </div>
<div class="gmail_default" style="font-family:trebuchet ms,sans-serif">Usando 24 como \lambda na expressão de V(X) tem-se que a variância é 48.</div>
<div class="gmail_default" style="font-family:trebuchet ms,sans-serif"> </div>
<div class="gmail_default" style="font-family:trebuchet ms,sans-serif">À disposição.<br>Walmes.</div>
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