Discrete random variables (1 Viewer)

leehuan · Jul 30, 2016

Hmm. I did think about the oldest possible age problem... the question wasn't specific in that regard.

I was trying to build an expression like that in the textbook but I couldn't figure out what to build.

InteGrand · Jul 30, 2016

InteGrand said:
I think you would need to make an assumption about what the oldest possible age is, e.g. 100 years (or call it N years or something), because that'd clearly affect the answer.

Also, Pr(X > 10 | X = 10) is 0, so I think you meant something else.

Part 2) is basically asking I think for E[X | X > 10], where X is the random variable ''age that person dies at''. This is known as a conditional expectation.

leehuan · Jul 30, 2016

InteGrand said:
Part 2) is basically asking I think for E[X | X > 10], where X is the random variable ''age that person dies at''. This is known as a conditional expectation.

Will research this later

InteGrand · Jul 30, 2016

Here's my guess (Q seemed a little vague).

$\noindent Since the number of people alive at each age is given to decrease in arithmetic progression, we will have to assume some maximum age. Let's just assume the maximum age is some positive integer $N$. So if a person gets to $N$ years old, then they die immediately.$

$\noindent If the number of people alive at each integer age decreases in arithmetic progression, then the probability distribution of $X$ (defined as before, i.e. ``age that person dies at'', which we will assume is integer-valued for simplicity) will be a \emph{discrete uniform distribution} on $\{1,2,\ldots, N-1,N\}$. This is because the condition about A.P. basically means that the proportion dying each year is the same, which makes the distribution uniform. (So note the probability mass function (pmf) of $X$ is $\mathbb{P}\left(X=k\right) \equiv p_k = \frac{1}{N}$ for $k=1,2,\ldots, N$.)$

$\noindent (It'd actually probably be nicer to make $X$ continuous and use a continuous uniform distribution, but we can't really do this unless we assume the A.P. condition occurs in-between integers years too, i.e. that the CDF is linear continuously. We are only actually old the CDF is linear in a discrete sense, so we don't actually know what happens inbetween the integers. But if you want you could assume it's linear there too, might make the problem nicer to deal with and more realistic, since people can live non-integer lifetimes.)$

$\noindent Now, if we have a 10-year old and we want the expected lifetime, this is just $\mathbb{E}\left[X\mid X \geq 11\right]$. We condition on $X\geq 11$ because being told we have a $10$-year old, so it must die at $11$ or later (assume $N>10$ of course). I also realised it wanted the future lifetime, so we need to subtract $10$ from this expectation, because $10$ years have already been lived and don't count to the `future' lifetime.$

$\noindent Now, note the conditional pmf of a discrete uniform (given $X\geq 11$, or any other integer in the support in fact) is discrete uniform (show as an exercise if you want, it's fairly intuitive though). Therefore, the \emph{conditional random variable} $X\mid X\geq 11$ (which is the value of $X$ conditional on the fact it is at least $11$) is uniform on $\left\{11,\ldots, N\right \}$. The mean of a uniform random variable is literally just the arithmetic mean of the possible values that it can take (easy exercise), so in this case, that is $\frac{11+N}{2}$. So $\mathbb{E}\left[X\mid X\geq 11\right]= \frac{11+N}{2}$. So the expected future lifetime would be $\frac{11+N}{2}-10$.$

$\noindent In summary: 1) A linearly decreasing number of people alive at each age (which means linear CDF) means the distribution of the r.v. $X$ is uniform;$

$\noindent 2) the conditional distribution of a uniform given $X\geq x_0+1$ is still uniform on the new interval $x_0$ to $N$;$

$\noindent 3) Therefore, the conditional expected age of death is $\mathbb{E}\left[X\mid X \geq x_0 +1\right] = \frac{x_0 +1+N}{2}$ (and subtract $x_0$ from this to find the \emph{future} lifetime). This summary is true for both the continuous and discrete uniform distributions, with slight modifications needed for the continuous case, namely, we wouldn't have $x_0+1$, just $x_0$, because we wouldn't need to jump to the next integer.$

leehuan · Aug 6, 2016

Two quickies. (Second one isn't really random I guess)

$Assume that the probability that the number of hailstorms occurring in a year has a Poisson distribution with $\lambda =\frac{1}{100}$. What is the probability that in a 100 year period there will be at least 2 years with only a single hailstorm?$

All I want to know about this one is that is this X~Poisson(1)?

$Assume that $n\text{ dice are tossed and the values on their uppermost faces, given by }X_1,X_2,\dots,X_n$ are arranged to give $Y=\frac{\sum_{i=1}^nX_i}{n}$. Determine $\\1. \mathbb{E}[X]\\ 2.Var[X]\\ $3. Comment on the differences between these answers and those in the previous exercise.$

In the previous exercise I got E[X]=7/2 and Var[X]=35/12 for just one die being rolled. I'm fairly sure that therefore E[Y]=7/2 and Var[Y]=35/(12n), but I'm not sure how to prove it.

InteGrand · Aug 6, 2016

leehuan said:
Two quickies. (Second one isn't really random I guess)

$Assume that the probability that the number of hailstorms occurring in a year has a Poisson distribution with $\lambda =\frac{1}{100}$. What is the probability that in a 100 year period there will be at least 2 years with only a single hailstorm?$

All I want to know about this one is that is this X~Poisson(1)?

$Assume that $n\text{ dice are tossed and the values on their uppermost faces, given by }X_1,X_2,\dots,X_n$ are arranged to give $Y=\frac{\sum_{i=1}^nX_i}{n}$. Determine $\\1. \mathbb{E}[X]\\ 2.Var[X]\\ $3. Comment on the differences between these answers and those in the previous exercise.$

In the previous exercise I got E[X]=7/2 and Var[X]=35/12 for just one die being rolled. I'm fairly sure that therefore E[Y]=7/2 and Var[Y]=35/(12n), but I'm not sure how to prove it.

Answer to your first Q: it's not Poisson(2).

For the dice one (note Y is the sample average):
1) The expectation of the sample average is just the expectation of each Xi, i.e. 3.5
2) The variance of the sample average is (sigma^2)/n, where sigma^2 is the variance of each Xi, which you can find as E[X^2] – (3.5)^2.

(In other words, what you thought were the answers are correct.)

These follow from basic facts about sample averages (due to properties of expectation and variance of independent r.v.'s).

leehuan · Aug 6, 2016

InteGrand said:
Answer to your first Q: it's not Poisson(2).

For the dice one (note Y is the sample average):
1) The expectation of the sample average is just the expectation of each Xi, i.e. 3.5
2) The variance of the sample average is (sigma^2)/n, where sigma^2 is the variance of each Xi, which you can find as E[X^2] – (3.5)^2.

(In other words, what you thought were the answers are correct.)

These follow from basic facts about sample averages (due to properties of expectation and variance of independent r.v.'s).

Yeah I edited back to 1 when I realised what I did lol. Lurking out of here though cause I've got a continuous r.v. qn now if you've got time

Discrete random variables (1 Viewer)

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