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2 years ago

Why is -a² is always negative (a ≠ 0) and why is (-a)² is always positive?

1 answer:

hjlf2 years ago

6 0

$-a^2=(-1)\cdot a\cdot a$

Regardless of the sign of $a$ , we have $a\cdot a=a^2\ge0$ (never negative). But multiplying by -1 makes it negative.

On the other hand,

$(-a)^2=((-1)\cdot a)^2=(-1)^2\cdot a^2=1\cdot a^2=a^2$

which can never be negative for real $a$ .

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Sliva [168]

Answer:

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Step-by-step explanation:

For this case we have a random sample $X_1 ,X_2,...,X_n$ where $X_i \sim N(\mu=\theta, \sigma)$ where $\sigma$ is fixed. And we want to show that the maximum likehood estimator for $\theta = \bar X$ .

The first step is obtain the probability distribution function for the random variable X. For this case each $X_i , i=1,...n$ have the following density function:

$f(x_i | \theta,\sigma^2) = \frac{1}{\sqrt{2\pi}\sigma} exp^{-\frac{(x-\theta)^2}{2\sigma^2}} , -\infty \leq x \leq \infty$

The likehood function is given by:

$L(\theta) = \prod_{i=1}^n f(x_i)$

Assuming independence between the random sample, and replacing the density function we have this:

$L(\theta) = (\frac{1}{\sqrt{2\pi \sigma^2}})^n exp (-\frac{1}{2\sigma^2} \sum_{i=1}^n (X_i-\theta)^2)$

Taking the natural log on btoh sides we got:

$l(\theta) = -\frac{n}{2} ln(\sqrt{2\pi\sigma^2}) - \frac{1}{2\sigma^2} \sum_{i=1}^n (X_i -\theta)^2$

Now if we take the derivate respect $\theta$ we will see this:

$l'(\theta) = \frac{1}{\sigma^2} \sum_{i=1}^n (X_i -\theta)$

And then the maximum occurs when $l'(\theta) = 0$ , and that is only satisfied if and only if:

$\hat \theta = \bar X$

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