If each of the observation $x_{1}, x_{2}, \ldots ., x_{n}$ is increased by $'a'$ where $a$ is a negative or positive number, show that the variance remains unchanged.
If each of the observation $x_{1}, x_{2}, \ldots ., x_{n}$ is increased by $'a'$ where $a$ is a negative or positive number, show that the variance remains unchanged.
Let $\bar{x}$ be the mean of $x_{1}, x_{2}, \ldots ., x_{n} .$ Then the variance is given by
$\sigma _1^2 = \frac{1}{n}\sum\limits_{i = 1}^n {{{\left( {{x_i} - \bar x} \right)}^2}} $
If $'a$ is added to each observation, the new observations will be
$y_{i}=x_{i}+a$ .......$(1)$
Let the mean of the new observations be $\bar{y} .$ Then
$\bar y = \frac{1}{n}\sum\limits_{i = 1}^n {{y_i} = \frac{1}{n}} \sum\limits_{i = 1}^n {\left( {{x_i} - a} \right)} $
$ = \frac{1}{n}\left[ {\sum\limits_{i = 1}^n {{x_i}} \sum\limits_{i = 1}^n a } \right] = \frac{1}{n}\sum\limits_{i = 1}^n {{x_i} + \frac{{na}}{n} = } \bar x + a$
i.e. $\bar{y}=\bar{x}+a$ ..........$(2)$
Thus, the variance of the new observations
$\sigma _2^2 = \frac{1}{n}\sum\limits_{i = 1}^n {{{\left( {{y_i} - \bar y} \right)}^2}} = \frac{1}{n}\sum\limits_{i = 1}^n {{{\left( {{x_i} + a - \bar x - a} \right)}^2}} $ [ Using $(1)$ and $(2)$ ]
$ = \frac{1}{n}\sum\limits_{i = 1}^n {{{\left( {{x_i} + \bar x} \right)}^2}} = \sigma _1^2$
Thus, the variance of the new observations is same as that of the original observations.
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