Exponentially distributed random variable

Question

Suppose that $X$ is an exponentially distributed random variable. If you know that $P(X>1)=a$ for some value $a$, then what is $P(1<X<4)$?

I really struggle with this question.

Answer 1

hope this helps: (let $X$ be the random variable you are talking about) \begin{align*} P(X>1) = 1 - P(X\leq 1) = 1 - F_{x}(1) = 1 - (1-e^{-\lambda 1}) = e^{-\lambda} \end{align*} And, \begin{align*} P(X>1) = a \ \Leftrightarrow e^{-\lambda} = a \Leftrightarrow \lambda = -\ln(a) \end{align*}

Once (by definition) $\lambda > 0$ we get that $a \in \hspace{0.1cm} ]0,1]$. And, more importantly, $X\sim Exp(-ln(a))$

To calculate $P(1<X<4)$ I suggest the following: \begin{align*} P(1<X<4) = F(4) - F(1) \end{align*}

Which can be calculated with two different procedures. (One is the same as the above, simply substituing the expression on the general formula for the cumulative Function)

Answer 2

Hint: for $X$ is distributed exponentially, $P(X>kx)=P(X>x)^k\quad\forall x,k>0$. In this case, try substituting $x=1,k=4$.

Answer 3

Memorylessness of the exponential distribution tells you $$\mathbb P(X> n+1 ) = \mathbb P(X >1) \mathbb P(X>n )= \mathbb P(X >1) ^n$$ for positive integer $n$. So $\mathbb P(X > 4 )= a^4$ and $$\mathbb P(1 < X \le 4) = \mathbb P(X >1) - \mathbb P(X >4) = a-a^4$$