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I am trying to find the MLE for a PDF with an unknown Distribution.

f(x;$\lambda$) = $\lambda^3x^5\ e^{-\lambda x^2}$

With bounds X $\ge$ 0.

The first thing I did was take the Big Pi notation of everything. Resulting in L($\lambda$) = $$\prod_{i=0}^n (\lambda^3x^5\ e^{-\lambda x^2})$$

I am confused now on how to proceed, because I cannot put a sigma in a an exponent and then take the Log of that.

Thank you.

Edit 2:

L($\lambda$) = $\sum_{i=0}^n x^5 \lambda^{3n}e^{-\lambda\sum_{i=0}^n x^2}$

I understand that this is the full expansion, what I am confused about is when I take the Log of the function as

Log(L($\lambda$)) = $Log(\sum_{i=0}^n x^5) 3nLog(\lambda)-\lambda\sum_{i=0}^n x^2$

I don't understand how to take the derivative afterwards.

The farthest I reached is

d/d$\lambda$ Log(L($\lambda$)) = $(3n/\lambda)$ and I don't understand how to continue from here.

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    $\begingroup$ You have nothing indexed inside the product. You should be computing $f(x_1, \dotsc, x_n; \lambda)$ for an observed sample of $n$ variates. $\endgroup$ – Nap D. Lover Dec 2 at 22:47
  • $\begingroup$ I edited the original question to add parenthesis around the product $\endgroup$ – air bmx Dec 2 at 23:10
  • $\begingroup$ That still does not fix what I was hinting at. Let me try to be more explicit. What you wrote is $\prod_{i=1}^n (\lambda^3 x^5 e^{-\lambda x^2})= (\lambda^3 x^5 e^{-\lambda x^2})^n$ but what I believe you are really after is $f(x_1,\dotsc, x_n ; \lambda)=(x_1 \cdot \dotsc \cdot x_n)^5 \lambda^{3n} e^{-\lambda(x_1^2+\dotsc + x_n^2)}$. Does this make sense? $\endgroup$ – Nap D. Lover Dec 2 at 23:20
  • $\begingroup$ Thank you for taking the time to answer my question. I see what you mean...I understand that I have to take the sum, but I am confused behind the math on simplifiying the equation to write it in terms of lambda hat after taking the log and then derivative. $\endgroup$ – air bmx Dec 2 at 23:55
  • $\begingroup$ I have written up an answer with most of the details carefully outlined. Please comment on it if you need more clarification. $\endgroup$ – Nap D. Lover Dec 3 at 0:33
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Let $X_1,\dotsc, X_n$ be an IID sample of RVs that all have density $f(x;\lambda)=\lambda^3 x^5 e^{-\lambda x^2}$ for $x>0$ and zero otherwise. We wish to find the MLE for $\lambda$. This can be done, typically, by maximizing the joint PDF $f(x_1,\dotsc, x_n;\lambda)$. By the IID assumption, we first have $$f(x_1,\dotsc, x_n;\lambda)=f(x_1;\lambda)\cdot \dotsc \cdot f(x_n;\lambda)$$ now substitute in the expression and simplify the products to get $$f(x_1,\dotsc, x_n; \lambda)=(\lambda^3 x_1^5 e^{-\lambda x_1^2}) \cdot \dotsc \cdot (\lambda^3 x_n^5 e^{-\lambda x_n^2})$$ $$=\lambda^{3n} (x_1\cdot \dotsc \cdot x_n)^5 e^{-\lambda (x_1^2+\dotsc +x_n^2)}.$$

Now maximizing this directly could be done, but often it is easier to maximize $\log f(x_1,\dotsc, x_n;\lambda)$, and the maximum of the latter is the same as the maximum of the former (do you know why?). Note that the variates $x_i$ are all $>0$ so all $\log x_i$ in the following are defined.

Thus, we take logarithms and using many properties of the logarithm ($\log(ab)=\log a+\log b$, that $\log (a^m)=m\log a$ and that $\log e^y=y$, for positive reals $a,b$, any real $m$ and any real $y$) to obtain $$l(\lambda):=\log f(x_1,\dotsc, x_n; \lambda) = 3n \log \lambda + 5\sum_{i=1}^n \log x_i-\lambda(x_1^2+\dotsc +x_n^2),$$ and finally we differentiate $l(\lambda)$ with respect to $\lambda$, set equal to zero to find critical points and check the one we find is a maximum by your favorite method from calculus. This results in $$l'(\lambda)=\frac{3n}{\lambda}-(x_1^2+\dotsc +x_n^2),$$ and thus for $l'(\lambda)=0$ we need $\hat{\lambda}=\frac{3n}{x_1^2+\dotsc +x_n^2}.$

Are all the steps clear?

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  • $\begingroup$ This makes 100% sense, thank you very much. I was struggling with the math, but I think I fully understand it now. $\endgroup$ – air bmx Dec 3 at 0:36

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