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While calculating P(2≤X≤4), for an exponential random distribution, the solution says, $P(2\leq X\leq 4) = F(4)-F(2)$, where F denotes the CDF.

My version is, P(2≤X≤4) = P(22) and P(X≤4), i.e. 1-P(X≤2) and P(X≤4) {1-F(2)} * F(4), presuming they are independent event.

I know it is wrong, but please clarify, where I am commiting mistake in this approach.

Thanks

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  • $\begingroup$ What is $P(22)$? The rest of the description is muddy. What events are independent? You need an equation $P(2\le X\le 4)=$. $\endgroup$ – herb steinberg Apr 17 at 16:48
  • $\begingroup$ The entire text is not coming once I post the question. $\endgroup$ – Pankaj Kumar Swain Apr 17 at 16:51
  • $\begingroup$ While calculating P(2≤X≤4), for an exponential random distribution, the solution says, P(2≤X≤4) = F(4)-F(2), where F denotes the CDF. My version is, P(2≤X≤4) = P(2<X≤4) as P(X=2) is zero for a continuous random variable. We can write, P(2<X≤4) = P(2<X) and P(X≤4), i.e. P(X>2) and P(X≤4), i.e. 1-P(X≤2) and P(X≤4) which equals {1-F(2)} * F(4), presuming 2<X and X≤4 are independent events. I know it is wrong, but please clarify, I am commiting mistake in this approach. Thanks $\endgroup$ – Pankaj Kumar Swain Apr 17 at 16:51
  • $\begingroup$ Very simply put, the events are not independent $\endgroup$ – Stan Tendijck Apr 17 at 17:03
  • $\begingroup$ Basic error: $X\gt 2$ and $X\le 4$ are NOT independent events, since they are both about the same random variable. I suggest you fix original statement - $P(22)$? $\endgroup$ – herb steinberg Apr 17 at 17:04

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