$X \sim Exp(\lambda)$ and it has a half life of $1$ (year) (lets say you have a set of x's and they decay and after 1 year you have half of the x's you had)


  1. Calculate the time when the expectancy $\mathbb{E}$ of the number of X's left is 10% of the number of X's at the beginning.

  2. If you have 1024 X's at the beginning, calculate the time when the expectancy of the number of X's is $1$

  3. The probability that no x from all the 1024 x's were gone after the time you calculate at #2

My Try:

  1. if the half life is $1$ then: $\frac{\ln{2}}{\lambda} = 1$ so: $\lambda = \ln{2}$
    $X \sim \text{Exp}(\ln{2})$ $f(x) = \ln(2) \cdot e^{-\ln(2)x} = \frac{\ln(2)}{2^x}$
    Now I need to find what time it takes for X to be 10% so $0.1$ of the original. the problem is that I don't know how much X's were in the first place. maybe 1 is the total percentage? (100%) ?

so: $\frac{1}{\frac{ln2}{2^x}} = 0.1 \Leftrightarrow 2^x = \ln(2) \cdot 0.1 \Leftrightarrow x = \log_{2}(\ln(2) \cdot 0.1) = $ ? I get negative result.. where am I wrong here?

  1. $1024 \cdot \ln(2) \cdot e^{-x} = 1$ and so the answer is $x = \ln(\ln(2) \cdot 1024) = 6.565$

  2. If we calculated the time where the expected number of x's is 1 then after infinite many years it would decay to completely $0$ no? I don't get this question...

Thank you very much!


1 Answer 1


An "exponentially distributed lifetime" means that $\mathsf E_t(X)=X_0\mathrm e^{-\lambda t}$ .. the expected amount of $X$ at time $t$ equals the original amount ($X_0$) times the decay coefficient.   We usually simply write $X(t)$ or $X_t$ for this expected amount.

The half-life, $t_{1/2}$, of this decay process is the time the expected amount equals half the original amount. So, $t_{1/2}=1\text{yr}$ indeed means, that $\tfrac 12=\mathrm e^{-\lambda (1\text{yr})}$, or $\lambda = \ln (2)\, \text{yr}^{-1}$ .

Similarly, you need to find $t$ such that $0.10=\mathrm e^{-\lambda t}$, or $$\begin{align}t ~&=~-\ln(0.10)/\ln(2)\\&\approx~3.32\end{align}$$


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