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I've tried to search for an answer to this question but did not manage to find one.

I have an assignment where we were given some data of a secret distribution, we are supposed to test these 3 models for the data.

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According to instructions we're supposed to use the point estimate of the mean & standard deviation to estimate the parameters a and b.

Now for my question: I've calculated the standard mean & deviation, but how can I estimate the parameters a and b using this information?

I don't see how I can use those to estimate a and b.

Thanks!

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  • $\begingroup$ First derive the expression for MLE estimates of $a$ and $b$. $\endgroup$ – Dhruv Kohli - expiTTp1z0 Nov 25 '16 at 13:34
  • $\begingroup$ Which point estimate? As @expiTT--I--1z0 says, MLE for $a$ and $b$ might be a good idea. $\endgroup$ – Shraddheya Shendre Nov 25 '16 at 14:14
  • $\begingroup$ There are some different possible answers to this question depending on your "fitting metric", but from context it sounds like they're actually intending for you to use a "method of moments estimator". That is, solve for the mean and variance as a function of $a,b$, set them equal to your sample mean and variance, then solve for $a,b$. $\endgroup$ – Ian Nov 25 '16 at 17:20
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  1. $f(x) = Ke^{-ax}, \, x \geq 0$

First, compute $K$ so that $f(x)$ is a probability distribution.

$$\int\limits_0^\infty Ke^{-ax} \, dx=1 \Rightarrow \frac{K}{a} = 1 \Rightarrow K = a$$

Likelihood expression will be,

$$L(a) = \prod_{i=1}^n f(x_i) = \prod_{i=1}^n ae^{-ax_i} = a^ne^{-a \sum\limits_{i=1}^n x_i}$$

$$\log(L(a)) = n\log(a) -a\sum_{i=1}^n x_i$$

Now, you would like to maximize the likelihood (which is equivalent to maximizing log likelihood) by computing the best possible value of $a$ (Maximum Likelihood Estimate (MLE) of $a$). Take the derivative of log likelihood and equate it to zero.

$$\frac{n}{a} - \sum_{i=1}^n x_i = 0 \Rightarrow a = \frac n {\sum\limits_{i=1}^n x_i}$$

Therefore, $a_\text{MLE} = \frac{n}{\sum\limits_{i=1}^n x_i} = \frac 1 {\hat{\mu}}$, where $\hat{\mu}$ is the sample mean.

  1. $f(x) = Kxe^{-ax}$

Find K.

$$\int\limits_0^\infty Kxe^{-ax} \, dx = 1 \Rightarrow 0 + \frac{K}{a}\int\limits_0^\infty e^{-ax} \, dx = 1 \Rightarrow \frac{K}{a^2} = 1 \Rightarrow K = a^2$$

Now, the expression of likelihood,

$$L(a) = \prod_{i=1}^n a^2x_i e^{-ax_i} = (a^2)^n \, e^{-a\sum\limits_{i=1}^n x_i} \prod_{i=1}^n x_i$$

$$\log(L(a)) = 2n\log(a) -a\sum\limits_{i=1}^n x_i + \sum_{i=1}^n \log(x_i)$$

Take derivative with respect to $a$ and equate to zero.

$$\frac{2n} a - \sum_{i=1}^{n}x_i = 0 \Rightarrow \frac 2 a - \hat{\mu} = 0 \Rightarrow a = \frac{\hat{\mu}} 2 $$

Now, I'll leave the third part to you.

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Presumably you have calculated the sample mean - usually called $\bar x$. This can be used as an unbiased estimate of the population mean. Let's call this value $\hat \mu$

Presumably you have also calculated the sample standard deviation. This is trickier. If you have used $n$ as the divisor, then what you have is a biased estimate of the standard deviation of the population. If you have used $n-1$ as the divisor, then what you have is an unbiased estimate of the population standard deviation. Let's call this $\hat \sigma$.

Now to consider the population mean $\mu$. This can be found from the pdf.

$\mu=\int_0^{\infty}xe^{-ax}dx$ - use integration by parts. This will give you an expression for $\mu$ in terms of $a$. You can equate this expression to $\hat \mu$ and rearrange to make $a$ the subject - there's your estimate of $a$.

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  • $\begingroup$ Thank you for the helpful answer! Somehow forgot you can calculate the mean by integrating the pdf. $\endgroup$ – user3284549 Nov 25 '16 at 15:05
  • $\begingroup$ Although the mean of the exponential distribution can be written in the form of the parameter $a$ but this might not be true for other distributions. The right way is to first compute the MLE of $a$ and $b$. $\endgroup$ – Dhruv Kohli - expiTTp1z0 Nov 25 '16 at 15:25
  • $\begingroup$ Minor point: with $n-1$ you have an unbiased estimate of the variance but not of the standard deviation. $\endgroup$ – Ian Nov 25 '16 at 17:17
  • $\begingroup$ @expiTT--I--1z0 You can definitely get the mean and variance explicitly, they are just the usual gamma function integrals with integer exponents. $\endgroup$ – Ian Nov 25 '16 at 22:33
  • $\begingroup$ Yes you are correct. $\endgroup$ – Dhruv Kohli - expiTTp1z0 Nov 25 '16 at 22:59

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