Answer to Question #130407 in Statistics and Probability for Himel abed

Question #130407
A random variable X has the probability density function given below

g(x)= {2x,0≤x<1
{ 6-x,1≤x<2
{ 0, otherwise

Show graphically and find out the probabilities: P(X≤3)
1
Expert's answer
2020-08-25T11:03:29-0400

g(x)={2x,0x<16x,1x<20,otherwiseg(x) = \begin{cases} 2x, & 0 \leq x<1 \\ 6-x, & 1\leq x <2 \\ 0, & otherwise\end{cases}

Plotting this, we get following



However, in order to be indeed probability density function condition +g(x)dx=1\int_{-\infty}^{+\infty} g(x)dx =1 must be fulfilled. In our case,

01(2x)dx+12(6x)dx=x201+6x12x2212=1+620.5=4.5.\int_0^1(2x)dx + \int_1^2(6-x)dx = x^2|_0^1 + 6x|_1^2 - \frac{x^2}{2}|_1^2 = 1 +6 -2 - 0.5 = 4.5.

In such case, g(x) should be normalized, namely g~(x)=g(x)/4.5\tilde{g}(x) = g(x)/4.5.

In order to find P(X3)P(X \leq3) we need to integrate g~(x)\tilde{g}(x) from -\infty to 3 or, what is the same, find the area below the (normalized) plot. Since for x3x \geq 3 g~(x)=0\tilde{g}(x) =0, integration from -\infty to 3 is the same as integration from -\infty to ++\infty. The last one we already did above.

P(X3)=3g(x)dx=3g~(x)dx4.5=+g~(x)dx4.5=1P(X \leq 3) = \int_{-\infty}^3 g(x)dx =\frac{\int_{-\infty}^3 \tilde{g}(x)dx }{4.5} = \frac{\int_{-\infty}^{+\infty}\tilde{g}(x)dx }{4.5} = 1.

Alternative approach: P(X3)=1P(X3)=1P(X \leq 3) = 1 - P(X \geq 3) = 1, since for x3x \geq 3 g(x)=0g(x) =0 and integral of it is 0 too.

Answer: P(X3)=1.P(X \leq 3) = 1.



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