Answer to Question #85788 in Statistics and Probability for Rohit Jain

Question

Which of the following statements are true or false?Give a short proof or a counter example in support of your answer.
(i)If the correlation coefficient between x and y is 0.75,then the correlation coefficient between (2+5x) and (-2y+3) is -0.75.
(ii)If P(A)=0.5,P(AUB)=0.7 and A and B are independent events,then P(B)=2/5.
(iii)Let X_1,X_2,...,X_n be a random sample of size n from N(0,σ^2).Then S^(2)_0=∑^n_i=1 X^(2)_i/σ^(2) follows normal distribution.
(iv)A maximum likelihood estimator is always unbiased.
(v)The mean deviation is least when deviations are taken about the mean.

Answer 1

(i) The correlation coefficient is:

$$r(x, y) = \frac{cov(x, y)}{\sqrt{\mathbb V(x) \mathbb V(y)}} = 0.75$$

Hence using the linearity of covariance:

$r(2 + 5 x, -2 y + 3) = \frac{cov(2 + 5 x, -2 y + 3)}{\sqrt{\mathbb V(2 + 5 x) \mathbb V(-2 y + 3)}} = \frac{-10 cov(x,y)}{25 \sqrt{\mathbb V(x) 4 \mathbb V(y)}} = \frac{-10}{10} 0.75 = -0.75$

So the first is true.

(ii) if P(A) = 0.5, P(A or B) = 0.7 and A and B are independent, then by Inclusion–exclusion principle:

$$P(A \cup B) = P(A) + P(B) - P(A \cap B) = P(A) + P(B) - P(A) P(B)$$ $$P(B) = \frac{P(A \cup B) - P(A)}{1 - P(A)} = \frac{0.2}{0.5} = 2/5$$

So the second is true.

(iii) Let's take n = 1, then

$$S = \frac{X_1 ^2}{\sigma ^2}$$

Since X_1 is N(0, sigma^2), S follows $N^2(0,1)$ . To see that it is not gaussian, one may ask what is the probability of S being less then zero. Clearly it is zero. But any normal distributed random variable would have non-zero probability of this event, hence the statement is wrong.

(iv) The simple counter-exemple is the following:

Consider the sample X_i of uniformly distributed random variables on the interval $(0,\theta)$.

The maximum likelihood estimator for theta would be

$$\bar \theta = \max_{1\leq i \leq n} X_i$$

To see this consider the product of densities:

$$L = \frac{1}{\theta} 1_{0 < X_1 < \theta} \cdot \frac{1}{\theta} 1_{0 < X_2 < \theta} ... \frac{1}{\theta} 1_{0 < X_n < \theta}$$

If $\bar \theta < max X_i$ then one of the indicators is 0 (the function $1_{X \in A}$ is called the indicator of the event A, it takes value 1 if X in A, 0 if not) and the whole L is 0.

if $\bar \theta \geq max X_i$ then

$$\frac{1}{\bar \theta^n} \geq \frac{1}{(\max_i X_i)^n}$$

but as mle is the argmax of L, so

$$\bar \theta = \max_i X_i$$

Now consider:

$$P(\bar \theta < \theta) > 0, P(\bar \theta \geq \theta) = 0$$

So clearly such an estimator cannot be unbiased.

(v) I assume you meant mean squared deviation. We want to prove that

$$\mathbb E X = \arg \min_F \sum_{i=1}^n (X_i - F)^2$$

Where

$$\mathbb E X = \frac{1}{n} \sum_{i=1}^n X_i$$

Let's take

$$\sum_{i=1}^n (X_i - F)^2 = \sum_{i=1}^n (X_i \pm \mathbb E X - F)^2 = \sum_{i=1}^n [(X_i - \mathbb E X)^2 + ( \mathbb E X - F)^2 + 2 (X_i - \mathbb E X) ( \mathbb E X - F)]$$

The last one here is 0:

$$\sum_i [2 (X_i - \mathbb E X) ( \mathbb E X - F)] =2 ( \mathbb E X - F) \sum_i (X_i - \frac{1}{n}\sum_j X_j)=2 ( \mathbb E X - F) (\sum_i X_i -\sum_j X_j) = 0$$

And we can rewrite:

$$\sum_{i=1}^n (X_i - \mathbb E X)^2 +n ( \mathbb E X - F)^2 \geq \sum_{i=1}^n (X_i - \mathbb E X)^2$$

Thus we see that $\sum_{i=1}^n (X_i - F)^2$ is greater than $\sum_{i=1}^n (X_i - \mathbb E X)^2$ for any choice of F.