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A) I calculate first the pdf. I'm not sure if it is correct.{I corrected the rages}

enter image description here

and from there I integrate the intervals given. My problem is that when I integrate the final cdf is not 1, clearly something wrong I'm doing.

C) I will use the pdf. (i) I integrate between - infitiny to +infinity.

*I need help with this step, I can not find the cdf that all the summ is
equal to 1.
[I break down all the interval and all the sum still gave me -5/2]

(c) Calculate P(X<3|X>2); (i) I integrate between -inifinity and 3. (ii) I integrate between 2 and infinity. and I obtain my finaly answer from I dividided (ii)/(i) Is that correct??

d) To find E(X) I will intagreate x*pfd in all the intervalls.

In conclusion, I want to know how to obtaing correctly cdf, and if my steps are correct. Thanks.

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  • 1
    $\begingroup$ The PDF is incorrect. Start with $[1,2]$, what is $f(2)$ already? $\endgroup$ – Did Sep 26 '15 at 21:25
  • $\begingroup$ Your $f(x)$ has value ?how much? when $x = 2.5$? Is this reasonable? $\endgroup$ – Dilip Sarwate Sep 26 '15 at 21:25
  • $\begingroup$ I corrected again. But I'm not sure yet. Thanks to be there :-) $\endgroup$ – JulietaR Sep 29 '15 at 18:54
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(A) Your PDF has a mistake: second line should be

$$ x/2 - 1/2\qquad\text{if $1\leq x\lt 2$}.$$

The CDF can then be found as follows:

$$F(x) = \begin{cases} x\leq 1 &: \quad 0 \\ & \\ 1\lt x\leq 2 &: \quad F(1) + \int_1^x(u/2-1/2)du = 0+\left[u^2/4-u/2\right]_1^x \\ & \quad = x^2/4-x/2-1/4 \\ & \\ 2\lt x\leq 3 &: \quad F(2) + \int_2^x(-u/2+3/2)du = 1/4+\left[-u^2/4+3u/2\right]_2^x \\ & \quad = -x^2/4+3x/2-7/4 \\ & \\ 3\lt x\leq 4 &: \quad F(3) = 1/2 \\ & \\ 4\lt x\leq 4.5 &: \quad F(4) + \int_4^x 1du = 1/2+\left[u\right]_4^x \\ & \quad = x-7/2 \\ & \\ 4.5\lt x &: \quad F(4.5) = 1. \end{cases}$$

(C) That's not quite right. The basic rule of conditional probability is, for events $A,B,\;$ that $P(A\mid B) = P(A\cap B)/P(B)$. Therefore,

\begin{align} P(X\lt 3\mid X\gt 2) &= \dfrac{P(2\lt X\lt 3)}{P(X\gt 2) } \\ &= \dfrac{F(3)-F(2)}{1-F(2)} \\ &= \dfrac{1/2-1/4}{1-1/4} = \dfrac{1}{3}. \end{align}

(D) Your method for $E(X)$ looks correct.

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