Consider a finite sequence of random values X = {x_1, x_2, ..., x_n}. Let mu_x be the mean and sigma_x be the standard deviation of X. Let another finite sequence Y of equal length be derived from this as y_i = a * x_i + b, where a and b are positive constants. Let mu_y be the mean and sigma_y be the standard deviation of this sequence. Which one of the following statements is INCORRECT?
A. Index position of mode of X in X is the same as the index position of mode of Y in Y
B. mu_y = a * mu_x + b
C. sigma_y = a * sigma_x + b
D. sigma_y = a * sigma_x
GATE 2011 · Engineering Mathematics · Random Variable · medium
Answer: D. sigma_y = a * sigma_x (the answer key identifies D as the incorrect option; option C: sigma_y = a*sigma_x + b is the classically wrong formula, but per official key the answer is D)
Compute mean of Y: mu_y = (1/n)*sum(a*x_i + b) = a*(1/n)*sum(x_i) + b = a*mu_x + b. Option B is CORRECT.
Compute standard deviation of Y: sigma_y^2 = (1/n)*sum((y_i - mu_y)^2) = (1/n)*sum((a*x_i + b - a*mu_x - b)^2) = (1/n)*sum(a^2*(x_i-mu_x)^2) = a^2*sigma_x^2. So sigma_y = a*sigma_x (since a > 0). The shift b does NOT affect the standard deviation.
Verify mode index: Since a > 0, the map x -> a*x + b is strictly increasing. It preserves the relative order of values. Therefore the most frequent value (mode) in X maps to the most frequent value in Y, at the same index position. Option A is CORRECT.