Suppose the random variable X has the following probability density function(PDF): open curly brackets table attributes columnalign left columnspacing 1.4ex end attributes row cell fraction numerator open vertical bar x close vertical bar over denominator 4 end fraction end cell cell negative 2 space less or equal than x less or equal than 2 end cell row 0 cell O t h e r w i s e end cell end table close If mu (X) = E(X) = 0, find Variance (X) ?
Question
Suppose the random variable X has the following probability density function(PDF): open curly brackets table attributes columnalign left columnspacing 1.4ex end attributes row cell fraction numerator open vertical bar x close vertical bar over denominator 4 end fraction end cell cell negative 2 space less or equal than x less or equal than 2 end cell row 0 cell O t h e r w i s e end cell end table close
If mu (X) = E(X) = 0, find Variance (X) ?
Solution
The variance of a random variable X, denoted as Var(X), is the expected value of the squared deviation from the mean. It is calculated as E[(X - μ)^2], where μ is the mean of X.
Given that the mean μ = E(X) = 0 for the random variable X, the variance can be simplified to E[X^2].
The probability density function (PDF) of X is given as |x|/4 for -2 ≤ x ≤ 2, and 0 otherwise.
To find the variance, we need to calculate E[X^2], which is the integral of x^2*f(x) dx from -2 to 2, where f(x) is the PDF.
So, E[X^2] = ∫ from -2 to 2 of (x^2 * |x|/4) dx.
This integral can be split into two parts due to the absolute value:
E[X^2] = ∫ from -2 to 0 of (x^2 * -x/4) dx + ∫ from 0 to 2 of (x^2 * x/4) dx = -1/4 * ∫ from -2 to 0 of x^3 dx + 1/4 * ∫ from 0 to 2 of x^3 dx = -1/4 * [(-2)^4/4 - 0] + 1/4 * [2^4/4 - 0] = -1/4 * 4 + 1/4 * 4 = 0.
So, the variance Var(X) = E[X^2] - (E[X])^2 = 0 - 0^2 = 0.
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