MHB V: Understanding the Calculation of Variance for a Given CDF

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The discussion revolves around calculating the variance of a random variable X defined by a specific cumulative distribution function (CDF). The key points include the calculation of the expected value E[X] using the integral of x multiplied by the derivative of the CDF, F'(x), for the interval 1 < x < 2. There is a consideration of the discrete nature of the variable at x=1, requiring the addition of 1*P[X=1] to the integral. The variance for regions where x < 0 and x ≥ 2 is stated to be zero, raising questions about their contribution to the overall variance calculation. The final clarification indicates that these regions are adequately represented in the integral, addressing concerns about missing components in the variance calculation.
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I already have the full solution to this so I'm not looking for an answer, I am hoping to get an explanation of certain parts that are unclear.

Question:
A random variable X has the following CDF:

[math] f(x) =\left\{ \begin{array}{lr} 0& x<0\\ \frac{x^2-2x+2}{2} & 1 \le x < 2 \\ 1 & x \ge 2 \end{array} \right.[/math]

Calculate the variance of X.

Solution:
When [math]1<x<2[/math] then E[X] can be calculated by first finding F'(x) and taking the integral [math]\int_{1}^{2}x*F'(x)dx[/math]

Since x must take on a particular value I'm guessing that it must be considered discrete at x=1 so to account for this we must add [math]1*P[X=1][/math] to the above integral.

For x < 0 and [math]x \ge 2[/math] the variance seems to be zero but I don't see why that isn't involved somehow in the calculation since x is apparently any value from negative infinity to infinity. The solution says the the above two parts plus finding the second moments of them are enough to find the variance but it seems to me a piece is missing.
 
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Houdini said:
I already have the full solution to this so I'm not looking for an answer, I am hoping to get an explanation of certain parts that are unclear.

Question:
A random variable X has the following CDF:

[math] f(x) =\left\{ \begin{array}{lr} 0& x<0\\ \frac{x^2-2x+2}{2} & 1 \le x < 2 \\ 1 & x \ge 2 \end{array} \right.[/math]

Calculate the variance of X.

Solution:
When [math]1<x<2[/math] then E[X] can be calculated by first finding F'(x) and taking the integral [math]\int_{1}^{2}x*F'(x)dx[/math]

Since x must take on a particular value I'm guessing that it must be considered discrete at x=1 so to account for this we must add [math]1*P[X=1][/math] to the above integral.

For x < 0 and [math]x \ge 2[/math] the variance seems to be zero but I don't see why that isn't involved somehow in the calculation since x is apparently any value from negative infinity to infinity. The solution says the the above two parts plus finding the second moments of them are enough to find the variance but it seems to me a piece is missing.

\(F'(x)=0\) for \(x<1\) and also for \(x > 2\) so these regions can be considered to already be represented in your integral of \(xF'(x)\)

CB
 
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