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How to compute the variance for the random variable Y that has the triangle distribution:

$$f_Y(x) = \begin{cases} 4x,\;0\leq x\ \leq \frac12 \\ 2-4x, \; \frac12 \leq x \leq 1 \end{cases} \quad$$ The answer should be $\frac{1}{24}$ but I don't know how to get this. This is what I have done: $$\mu_Y = \int_{\frac12}^{0} 4x\cdot x \,dx +\int_{1}^{\frac12} (2-4x)\cdot x \,dx = -0.25$$ $$\text{Var}(Y) = E(Y^2) - \mu_Y^2 = \int_{\frac12}^{0} 4x\cdot (x+0.25)^2 \,dx +\int_{1}^{\frac12} (2-4x)\cdot (x+0.25)^2 \,dx$$ I could not get the correct answer and did not have a clear idea what I've done incorrectly. Any help will be appraciated.

LavO
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  • You are using a non-standard order of integration borders (higher number should stand high), therefore, while exchanging the integral with the antiderivative values, you may have gotten false signs. – user7427029 Nov 15 '21 at 18:02
  • And how did you came up with $(x + 0,25)^2$? Why not only $x^2$ (law of unconscious statistician)? – user7427029 Nov 15 '21 at 18:04
  • Definition of $f_Y(x)$ is bad. $2-4x \le 0$ for upper half. – herb steinberg Nov 15 '21 at 19:04

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