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Given $$f ( x_1 , x_2 ) = \begin{cases}6 ( 1 - x_2 ) &,&0 ≤ x_1 ≤ x_2 ≤ 1 \\ 0 &&\text{ Otherwise.}\end{cases}$$ Find $P ( X_1 ≤ 0.75, X_2 ≥ 0.5 )$

The correct answer is $31/64$.

Help me know the lower and upper limit to use in differentiating the function. I used $0$ and $3/4$ for $x_2$ and $0.5$ and $1$ for $x_1$ but got $18/32$.

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    $\begingroup$ Have you taken into account that integrand $f(x_1,x_2)$ takes value $0$ if $x_2<x_1$? $\endgroup$ – drhab Jul 17 '18 at 11:19
  • $\begingroup$ Also you have the bounds backwards. $\endgroup$ – Graham Kemp Jul 17 '18 at 13:53
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$$P ( X_1 ≤ 0.75, X_2 ≥ 0.5 )=1-P ( X_1 ≤ 0.75, X2 \le 0.5 )$$also $$P ( X_1 ≤ 0.75, X_2 \le 0.5 )=\int_{0}^{0.75}\int_{x_1}^{0.5}6-6x_2dx_2dx_1=\int_{0}^{0.75}2.25-6x_1+3x_1^2dx_1=\dfrac{33}{64}$$therefore $$P ( X_1 ≤ 0.75, X_2 ≥ 0.5 )=1-P ( X_1 ≤ 0.75, X2 \le 0.5 )=\dfrac{31}{64}$$

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  • $\begingroup$ for this part P(X1≤0.75,X2≤0.5) I don't get 33/64, instead I get 27/64. @Mostafa Ayaz $\endgroup$ – Eric kioko Jul 17 '18 at 11:53
  • $\begingroup$ Also $\mathsf P(X_1\leqslant a, X_2\geqslant b)= \mathsf P(X_1\leqslant a)-\mathsf P(X_1\leqslant a, X_2< b)$ $\endgroup$ – Graham Kemp Jul 17 '18 at 13:55
  • $\begingroup$ So use instead:$$\begin{align}\mathsf P(X_1\leqslant 3/4)&=\int_0^{3/4}\int_x^1 6(1-y)\mathsf d y\mathsf d x\\[2ex]\mathsf P(X_1\leqslant 3/4, X_2<1/2)&=\int_0^{1/2}\int_x^{1/2} 6(1-y)\mathsf d y\mathsf d x &\text{since }X_1\leqslant X_2\end{align}$$ $\endgroup$ – Graham Kemp Jul 17 '18 at 14:24
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The key is to stay within your support. $$\begin{align}\mathsf P(X_1\leqslant 3/4, X_2\geqslant 1/2) &= \int_{-\infty}^{3/4}\int_{1/2}^\infty 6(1-x_2)\mathbf 1_{0\leqslant x_1\leqslant x_2\leqslant 1}~\mathsf dx_2~\mathsf dx_1\\[1ex] &=\int_0^{3/4}\!\!\int_{\max\{x_1,1/2\}}^1 6(1-x_2)~\mathsf dx_2~\mathsf dx_1\\[1ex] &~~\vdots\\[1ex] &=\dfrac{31}{64}\end{align}$$


PS: use the obvious fact that $\max\{1/2,x_1\}=\begin{cases} 1/2&:& x_1<1/2\\x_1&:& 1/2\leqslant x_1\end{cases}$ to partition the integral into a sum of two integrals.

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  • $\begingroup$ The problem is what is the max or rather how to get the max here,if you could shed some more light. $\endgroup$ – Eric kioko Jul 17 '18 at 19:54
  • $\begingroup$ $x_1\leqslant x_2$ and $1/2\leqslant x_2$ means $\max\{x_1,1/2\}\leqslant x_2$ @Erickioko $\endgroup$ – Graham Kemp Jul 17 '18 at 21:21

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