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Question #41826

Difficulty: 8/10 | 🟢 Free

The Question

Let a discrete random variable $X$ take values in the set $S = \{-1, 0, 1\}$ . For a real parameter $t$ , the probability mass function of $X$ is given by $P(X = x) = c_t (4 - (x-t)^2), \quad x \in S$ where $c_t > 0$ is a normalizing constant depending on $t$ . Let $M$ and $m$ be the maximum and minimum possible values of the variance of $X$ over all valid parameters $t$ for which $P(X=x)$ is a well-defined probability distribution. Find the value of $M + m$ .

Options

A.
$\frac{207}{245}$
B.
$\frac{6}{5}$
C.
$\frac{12}{49}$
D.
$\frac{213}{245}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Probability Mass Function (PMF)|Variance Invariance under Translation|Concavity Analysis

Why This Question Is Tricky

Common Mistakes

Students often incorrectly assume that translating the parameter $t$ in the PMF $f(x-t)$ simply translates the distribution without changing its variance (i.e., assuming $Var(X+c) = Var(X)$ applies to the parameter shift).

However, because the support is fixed at $S=\{-1, 0, 1\}$ , changing $t$ alters the shape of the distribution and the normalizing constant $c_t$ .

The trap lies in failing to identify the valid domain of $t$ ( $t \in [-1, 1]$ ) required to keep $P(X=x) \ge 0$ .

By applying concavity analysis (AOD) on the variance function $V(t^2)$ , students must deduce that the variance is strictly decreasing with respect to $t^2$ , yielding the extrema exactly at the boundaries of the valid parameter space.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Probability - Class 12
Concept	Probability Mass Function (PMF)
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free