Burning time of Christmas Tree

[mathmari]

Hey! :giggle:

The burning time of an electric candle is between $60$ and $80$ hours. It is considered to be continuously uniformly distributed over this period. A Christmas tree with $12$ candles (independent of the burning time), only lights up as long as all candles are working.
Determine, with intermediate steps, for the "burn time" of the tree, the distribution function and the average burn duration.
Hint : To do this, model the burning time of the individual candles using suitable independent and identically distributed random variables $X_1, X_2,\ldots , X_{12}$. The burn duration of the tree is then given by $Y = \min \{X_1, X_2,\ldots , X_{12}\}$. First determine $P [Y \geq x]$ and note that the minimum of $12$ real numbers is greater than or equal to $x$ when all $12$ numbers are greater than or equal to $x$.

Could you explain to me why $Y$ is defined as the minimum of all $X_i$'s ?
We have to calculate the distribution function $F_Y(y)$ and the average burn duration, i.e. the expected value $E[Y]$, right?

:unsure:

 

[I like Serena]

Hey mathmari!

The minimum identifies the candle that stops working first. Once it stops, all candles stop. So the burning time Y of the tree is equal to that minimum.

First we have to identify the distribution functions of $X_i$, and then we should indeed find $f_Y$ and $EY$.

 

[mathmari]

The minimum identifies the candle that stops working first. Once it stops, all candles stop. So the burning time Y of the tree is equal to that minimum.

First we have to identify the distribution functions of $X_i$, and then we should indeed find $f_Y$ and $EY$.

So $X_i$ describes the burning time of each candle ? The burn time of the tree is the time till the first candle stops burning, i.e. it is eual to the minimum of all the burning times of the candles, so it is described by $Y=\min \{X_1, \ldots , X_{12}\}$, right? :unsure:

Why do we have to determine $P[Y\geq x]$ ? If it is to calculate $F_Y(x)$ then we need $P[Y\leq x]$, or not? :unsure:

What does it mean that the burning time of an electric candle is between $60$ and $80$ hours ? Does itmean that $X_i\sim U([60,80])$ ? :unsure:

 

[I like Serena]

So $X_i$ describes the burning time of each candle ? The burn time of the tree is the time till the first candle stops burning, i.e. it is eual to the minimum of all the burning times of the candles, so it is described by $Y=\min \{X_1, \ldots , X_{12}\}$, right?

Why do we have to determine $P[Y\geq x]$ ? If it is to calculate $F_Y(x)$ then we need $P[Y\leq x]$, or not?

What does it mean that the burning time of an electric candle is between $60$ and $80$ hours ? Does itmean that $X_i\sim U([60,80])$ ?

Yes to all. (Nod)
In particular it also says "continuously uniformly distributed", so indeed $X_i \sim U$.

It's a hint to determine $P[Y\geq x]$. And yes we would need $P[Y\leq x]$, which is equal to $1-P[Y\geq x]$.

 

[mathmari]

Yes to all. (Nod)
In particular it also says "continuously uniformly distributed", so indeed $X_i \sim U$.

It's a hint to determine $P[Y\geq x]$. And yes we would need $P[Y\leq x]$, which is equal to $1-P[Y\geq x]$.

So we have the following :
\begin{equation*}F_Y(x)=P[Y\leq x]=1-P[Y>x]=1-P[\min \{X_1, \ldots , X_{12}\}>x]=1-P[\{X_1>x\}\cap \{X_2>x\}\cap \ldots \cap \{X_{12}>x\}]\end{equation*} Is that correct so far? Or do we use $\cup$ instead of $\cap$ ? Or should it be with commas like at joint distributions? :unsure:

 

[mathmari]

Do we maybe have the following ?
\begin{align*}F_Y(x)&=P[Y\leq x]\\ & =1-P[Y>x]\\ & =1-P[\min \{X_1, \ldots , X_{12}\}>x]\\ & =1-P[X_1>x, \ X_2>x, \ldots , \ X_{12}>x]\\ & \ \overset{X_i\text{ independent }}{=}\ 1-P[X_1>x]\cdot P[X_2>x]\cdot \ldots \cdot P[X_{12}>x]\\ & =1-\left (1-P[X_1\leq x]\right )\cdot \left (1-P[X_2\leq x]\right )\cdot \ldots \cdot \left (1-P[X_{12}\leq x]\right )\\ & =1-\left (1-F(x)\right )\cdot \left (1-F(x)\right )\cdot \ldots \cdot \left (1-F(x)\right )\\ & =1-\left (1-F(x)\right )^{12}\\ & =\begin{cases}1-\left (1-0\right )^{12} & \text { if } x<60 \\ 1-\left (1-\frac{x-60}{80-60}\right )^{12} & \text { if } 60\leq x\leq 80 \\ 1-\left (1-1\right )^{12} & \text { if } x>80\end{cases}\\ & =\begin{cases}0 & \text { if } x<60 \\ 1-\left (1-\frac{x-60}{20}\right )^{12} & \text { if } 60\leq x\leq 80 \\ 1 & \text { if } x>80\end{cases} \end{align*}
:unsure:

 

[mathmari]

The average burn duration , i.e. the expected value of $Y$ is equal to \begin{equation*}E(Y) = \int_{-\infty}^{\infty} y f_Y(y) \, dy\end{equation*}
Thedensity is the derivative of the distribution function \begin{align*}f_Y(y)&=F_Y'(y)=\begin{cases}0 & \text { if } y<60 \\ -12\left (1-\frac{y-60}{20}\right )^{11}\cdot \left (-\frac{1}{20}\right ) & \text { if } 60\leq y\leq 80 \\ 0 & \text { if } y>80\end{cases}=\begin{cases}0 & \text { if } y<60 \text{ or } y>80\\ \frac{12}{20}\left (1-\frac{y-60}{20}\right )^{11} & \text { if } 60\leq y\leq 80 \end{cases}\\ & =\begin{cases}0 & \text { if } y<60 \text{ or } y>80\\ \frac{3}{5}\left (1-\frac{y-60}{20}\right )^{11} & \text { if } 60\leq y\leq 80 \end{cases}\end{align*}
Therefore \begin{align*}E(Y)& = \int_{-\infty}^{\infty} y f_Y(y) \, dy\\ & =\int_{-\infty}^{60} y f_Y(y) \, dy+\int_{60}^{80} y f_Y(y) \, dy+\int_{80}^{\infty} y f_Y(y) \, dy\\ & =\int_{-\infty}^{60} y \cdot 0 \, dy+\int_{60}^{80} y \cdot \frac{3}{5}\left (1-\frac{y-60}{20}\right )^{11} \, dy+\int_{80}^{\infty} y \cdot 0 \, dy\\ & =\frac{3}{5}\cdot \int_{60}^{80} y \cdot \left (1-\frac{y-60}{20}\right )^{11} \, dy\\ & =\frac{3}{5}\cdot \frac{4000}{39}= \frac{800}{13}\approx 61.5 \text{ hours }\end{align*} Is that correct? :unsure:

 

[I like Serena]

It all looks correct to me. (Sun)

And I think we can use either commas inside the set of outcomes, or $\cap$ symbols between the sets of outcomes.

 

[mathmari]

It all looks correct to me. (Sun)

And I think we can use either commas inside the set of outcomes, or $\cap$ symbols between the sets of outcomes.

Great! Thank you very much! (Sun)