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Let's assume that X is a random variable which follows exponential distribution. The expected value of this distribution is E . How can I compute the following probability:

$P(X\leq E)$

Thank you so much

Angelıque
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  • What are your thoughts? – StubbornAtom Jul 06 '18 at 10:01
  • $E$ is a constant so can only be looked at as degenerated random variable in this context. You write $P(E\mid X\leq E)$ but I cannot recognize a conditional probability in that. This because in $E$ I do not recognize an event. – Vera Jul 06 '18 at 10:05
  • Did you possibly mean $P(X,|,X≤E)$? – lulu Jul 06 '18 at 10:08
  • @lulu And even if so. What is the meaning of $P(X\mid X\leq E)$? Is it some way to denote the conditional distribution of $X$ (under $X\leq E$)? I am not familiar with that notation and it makes a confusing impression on me. – Vera Jul 06 '18 at 10:11
  • @Vera Yes, at least that's how I would read that, as $P(X=x,|,X≤E)$. Or the OP might intend to ask for the expected value of $X$ given that $X≤E$. It sure isn't clear what is intended here. – lulu Jul 06 '18 at 10:18
  • @lulu you are right , i am asking the expected value of X given that X≤E , thank you – Angelıque Jul 06 '18 at 10:39
  • Please edit your post to reflect that. For the question: the probability that $X=x$ conditioned on $X≤E$ is $0$ if $x>E$ and otherwise $\frac {P(X=x)}{P(X≤E)}$. So, first you will have to compute $P(X≤E)$. – lulu Jul 06 '18 at 10:43
  • I edited it , thank you. The $P(X\leq E) $does not correspond the standard deviation which is $1/\lambda$ since E is the expected value of exponential distribution , I am confused at that part – Angelıque Jul 06 '18 at 10:53
  • Not following, If a random variable $Z$ has a density function $P_Z(z)$ then the probability that $Z≤A$ is $\int_{-\infty}^A P_Z(z),dz$. – lulu Jul 06 '18 at 10:59
  • Yes, I had thought the same thing , let me explain how I did: X is a random variable wich follows expo. dist. so the density function is $e^{-\lambda.x}$ . So my integration will be $\int_{0}^{E} e^{-\lambda.x} dx$ . When I compute this integration , I get $(1/\lambda)-(1/e)$ . I do not know if I am correct – Angelıque Jul 06 '18 at 11:25
  • The density for the exponential distribution is $\lambda,e^{-\lambda x}$. Your integral is close but not correct. – lulu Jul 06 '18 at 11:31
  • Ah , yes , then the result will be $1-(1/e)$, which I am not sure – Angelıque Jul 06 '18 at 11:40
  • Yes, that's the answer. – lulu Jul 06 '18 at 11:42
  • Then can we interpret this expression like that: Independent from what is the expected value of distribution (which will change according to $\lambda$ the probability of having a random value less than the expected value is always the same which is 1-(1/e) ? – Angelıque Jul 06 '18 at 11:48
  • Yes, that is correct. – lulu Jul 06 '18 at 12:19

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The probability distribution function $P(X\le x)=1-e^{-kx}$. The density function $f(x)=ke^{-kx}$ The mean $E(X)=\int_0^{\infty}kxe^{-kx}dx=\frac{1}{k}$. You have $E=\frac{1}{k}$. Therefore $P(X\le E)=1-e^{-1}$.

The title mentions conditional probability. Where is it?

herb steinberg
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