Finding variance of random variables that are exponentially distributed

Question

Suppose that lifetime of bulb of type n is exponentially distributed with mean value n years, where $n = 1, 2, 3, 4, 5$.

A bulb is equally likely be chosen from these five types.

(a) Find the probability that the selected bulb can survive for at least 3 years.

(b) Find the variance of the lifetime of the selected bulb.

For (a) I split the cases into 5 cases and calculated the $1-P(X<2)$ which equated to $e^{2/n}$ which led to $\sum_{i=1}^{5}0.2 * e^{2/i}$

For (b) I'm not sure how to start, am i supposed to use $Var(X) = 0.2Var(X_1) + 0.2Var(X_2) + ... + 0.2Var(X_5)$ since they have independent lifetimes.

Answer 1

If you do not want to use the law of total variance, which would give $\frac{1+4+9+16+25}{5} + 2$, then an alternative approach here is:

• The mean of $X_n$ is $n$ and the variance of $X_n$ is $n^2$, so $E[X_n^2]=2n^2$.

• You can average the means for your mixture distribution so $E[X]=\frac{1+2+3+4+5}5=3$.

• You cannot simply average the variances but you can average the $E[X_n^2]$ to get $E[X^2]=\frac{2+8+18+32+50}{5}=22$.

• That gives $\text{Var}(X)=E[X^2]-(E[X])^2=22-9=13$ for your mixture distribution.

Answer 2

Let $N$ be the uniform random variable indicating which among the five types is selected. $$N\sim\mathcal U\{1,2,3,4,5\}\\X\mid N\sim\mathcal{Exp}(1/N)$$

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(a) Exponential distributions are continuous, So the complement of lasting at least 3 years is lasting less than 3 years, rather than lasting less than 2 years.

$\qquad\begin{align}\mathsf P(X\geqslant 3) &= \mathsf E\big(1-\mathsf P(X\lt 3\mid N)\big)\\&= \mathsf E(\mathrm e^{-3/N})\\ &= \tfrac 15\sum_{n=0}^5\mathrm e^{-3/n}\\&~~\vdots\end{align}$

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(b) Use the Law of Total Variance.

$\qquad\begin{align}\mathsf{Var}(X) &=\mathsf E(\mathsf{Var}(X\mid N))+ \mathsf {Var}(\mathsf E(X\mid N))\\ &=\mathsf E(N^2)+\mathsf{Var}(N)\\ &~~\vdots\end{align}$