# Is $P\{X_1\leq \alpha- X_2\leq X_2\}=F_{X_1}(\alpha)\left[1-F_{X_2}(\frac{\alpha}{2})\right]$?.

Let $$X_1$$ and $$X_2$$ two independent random variables with CDF $$F_{X_1}(x_1)$$ and $$F_{X_2}(x_2)$$.

Can we say or it is true that $$P\{X_1\leq \alpha- X_2\leq X_2\}=P\{X\leq \alpha\leq 2X_2\}$$

$$P\{X_1\leq \alpha\leq 2X_2\}=F_{X_1}(\alpha)\left[1-F_{X_2}(\frac{\alpha}{2})\right]$$.

Thanks

• What is $X$? You are given $X_1$ and $X_2$ but there is no $X$ here. – Kavi Rama Murthy Jun 14 at 23:36
• Hi, yes I am sorry I forget it is $X_1$ – Monir Jun 14 at 23:38
• hi I am try to avoiding something like that $F_{x_i}(\alpha-x_i).$ – Monir Jun 14 at 23:40
• because in my PDFs and CDFs, for example if I use $\int_{x_1=0}^{\alpha}f_{X_1}(x_1)F_{X_2}(\alpha-x_1)$ – Monir Jun 14 at 23:43
• You cannot use densities unless you are told that the given random variables have densities – Kavi Rama Murthy Jun 14 at 23:46

Both are wrong . For the first one $$X_1=1,X_2=2$$ and $$\alpha =2$$ gives a counterexample. For the second one you need continuity of $$F_{X_2}$$ at the point $$\frac {\alpha} 2$$. In general you should take the left hand limit of $$F_{X_2}$$ at the point $$\frac {\alpha} 2$$ on RHS.

• So it probability zero? – Monir Jun 14 at 23:45
• Yes, in my example LHS is $0$ but RHS is $1$. – Kavi Rama Murthy Jun 14 at 23:47
• So there cases where is not zero or it alwas zeros ($X_1\leq \alpha-X_2$ and $\alpha-X_2\leq X_2$) so we can not express using CDF,? – Monir Jun 14 at 23:49
• $X_1 \leq \alpha -X_2$ means $X_1+X_2 \leq \alpha$ so you have to carry out an an integration over a suiatble subset of $\mathbb R^{2}$ w.r.t. the joint distribution. – Kavi Rama Murthy Jun 14 at 23:52
• Can I modifies this question to put all my problem or I add new question, because first I was find diffuct to slove this probability $P\{X_1+X_2\leq \alpha, X_1\leq X_2\}$. – Monir Jun 14 at 23:54