ACTL1101 Questions Help (mostly first year uni probability) (1 Viewer)

1008 · Aug 7, 2016

Hi

I am doing ACTL1101 at UNSW this semester, so I'd like to ask a few first year probability questions here:

$\\\text{Suppose that P(X=0)=1 - P(X=1). If E(X) = 3Var(X), find P(X=0).}$

InteGrand · Aug 8, 2016

1008 said:
Hi

I am doing ACTL1101 at UNSW this semester, so I'd like to ask a few first year probability questions here:

$\\\text{Suppose that P(X=0)=1 - P(X=1). If E(X) = 3Var(X), find P(X=0).}$

$\noindent Let $p\equiv \mathbb{P}\left(X=1\right)$. Then $X$ is a $\text{Bernoulli}\left(p\right)$ random variable. Using standard formulas for mean and variance of a Bernoulli r.v., we have $\mu_X \equiv \mathbb{E}\left[X\right] = p$ and $\sigma^2 _X \equiv \mathrm{Var}\left[X\right] = pq$, where $q \equiv 1-p=\mathbb{P}\left(X =0\right)$. We are thus given that $p = 3pq$. So either $p=0 \Rightarrow q = 1$ or $p\neq 0$, in which case we can cancel $p$ to find $q = \frac{1}{3}$. I.e. The probability in question is either $1$ or $\frac{1}{3}$.$

1008 · Aug 8, 2016

Thanks

$\noindent A certain community is composed of $m$ families, $n_i$ of which have $i$ children, $ \sum_{n=i}^{r}n_i=m$. If one of the families is randomly chosen, let $X$ denote the number of children in that family. If one of the $ \sum_{n=i}^{r}in_i$ children is randomly chosen, let $Y$ denote the total number of children in the family of that child. Show that $E[Y]\geq E[X]$

InteGrand · Aug 8, 2016

1008 said:
Thanks

$\noindent A certain community is composed of $m$ families, $n_i$ of which have $i$ children, $ \sum_{n=i}^{r}n_i=m$. If one of the families is randomly chosen, let $X$ denote the number of children in that family. If one of the $ \sum_{n=i}^{r}in_i$ children is randomly chosen, let $Y$ denote the total number of children in the family of that child. Show that $E[Y]\geq E[X]$

I think there are typos here because the sums don't really make sense. (I presume i was meant to be a summation index, but it appears in the limits of summation too, which a summation index should never do. Also, r hasn't been defined I think.)

I guess the first sum should just say that the sum over all i of n_i is m. (I.e. Total no. of families is m.) In fact, for both sums, the sums are just over all values of i.

1008 · Aug 8, 2016

InteGrand said:
I think there are typos here because the sums don't really make sense. (I presume i was meant to be a summation index, but it appears in the limits of summation too, which a summation index should never do. Also, r hasn't been defined I think.)

I guess the first sum should just say that the sum over all i of n_i is m. (I.e. Total no. of families is m.)

I checked but I couldn't find any errors/typos I've made. I think you can assume r to be the highest number of children in one single family.

InteGrand · Aug 8, 2016

1008 said:
I checked but I couldn't find any errors/typos I've made. I think you can assume r to be the highest number of children in one single family.

Don't worry, I think I've figured out what the Q. is asking. If you haven't made a typo, then the Q. itself essentially has a typo I think, because i is meant to be the thing we're summing over, yet it appears as the lower limit of a sum, which cannot happen.

InteGrand · Aug 8, 2016

Here are some hints for a method:

• For any given i, find an expression for Pr(X = i), i.e. The probability that the randomly chosen family is a family with i children. This expression is very simple and easy to find once you understand the notation used.

• Thus, write down an expression for E[X] using the definition of expectation for discrete r.v.'s.

• Repeat the above two steps for Y (again, this is very simple to do once you've gotten your head around the notation).

• Using your expressions for E[Y] and E[X], derive the required inequality (it shouldn't be too hard to do once you have the expressions).

1008 · Aug 8, 2016

InteGrand said:
Here are some hints for a method:

• For any given i, find an expression for Pr(X = i), i.e. The probability that the randomly chosen family is a family with i children. This expression is very simple and easy to find once you understand the notation used.

• Thus, write down an expression for E[X] using the definition of expectation for discrete r.v.'s.

• Repeat the above two steps for Y (again, this is very simple to do once you've gotten your head around the notation).

• Using your expressions for E[Y] and E[X], derive the required inequality (it shouldn't be too hard to do once you have the expressions).

Alright thanks for that, will look into it tomorrow and will let you know how I went.

leehuan · Aug 8, 2016

Where are these questions from?

1008 · Aug 8, 2016

leehuan said:
Where are these questions from?

One of my friends asked me...I have no idea where he got it from...

1008 · Aug 8, 2016

InteGrand said:
Here are some hints for a method:

• For any given i, find an expression for Pr(X = i), i.e. The probability that the randomly chosen family is a family with i children. This expression is very simple and easy to find once you understand the notation used.

• Thus, write down an expression for E[X] using the definition of expectation for discrete r.v.'s.

• Repeat the above two steps for Y (again, this is very simple to do once you've gotten your head around the notation).

• Using your expressions for E[Y] and E[X], derive the required inequality (it shouldn't be too hard to do once you have the expressions).

Oh and yeah thanks I figured it out...

leehuan · Aug 8, 2016

Lols

Tbh whilst I'm here, it's like the easiest distribution to learn (just a special case of binomial) but I don't think the Bernoulli distribution explicitly is a part of ACTL1101

leehuan · Aug 28, 2016

Something's weird...

Just by inspection I could tell that the expectation for the second case would be higher than the first. But why should I not be surprised that the variances are the exact same?

I have a feeling this isn't just by coincidence...

InteGrand · Aug 28, 2016

leehuan said:
Something's weird...

Just by inspection I could tell that the expectation for the second case would be higher than the first. But why should I not be surprised that the variances are the exact same?

I have a feeling this isn't just by coincidence...

Note that A = 25000 – B, where A is the r.v. describing the value from investment A, and similarly for B.

Therefore, by properties of variance (namely that an additive constant doesn't change the variance, and neither does the presence of a multiplicative factor of -1), we have

Var(A) = Var(25000 – B) = Var(B).

1008 · Sep 3, 2016

Is it true that:

$\text{If }X_1 \sim \text{Pois}(\lambda_1), X_2 \sim \text{Pois}(\lambda_2)\text{ then }(X_1+X_2=n)\sim\text{Pois}(\lambda_1+ \lambda_2)$

If so, what theorem is it?

InteGrand · Sep 3, 2016

1008 said:
Is it true that:

$\text{If }X_1 \sim \text{Pois}(\lambda_1), X_2 \sim \text{Pois}(\lambda_2)\text{ then }(X_1+X_2=n)\sim\text{Pois}(\lambda_1+ \lambda_2)$

If so, what theorem is it?

Yeah that's a true statement (assuming those are independent r.v.'s). That says that a sum of independent Poisson random variables is Poisson with rate parameter equal to the sum of the original rate parameters. It was proved by leehuan as an old homework exercise I believe (but the proof isn't written up here): http://community.boredofstudies.org...theorem-normal-distributions.html#post7161430 .

InteGrand · Sep 3, 2016

If you want to actually see a proof, you can find one here: http://math.stackexchange.com/quest...f-sum-of-two-random-independent-variables-x-y .

leehuan · Sep 3, 2016

Confession: Very screwed for this course. lol

leehuan · Sep 3, 2016

Please explain where their product came from?

Company NewGen has one electirc generator that has five components. All five components are required to work to produce electricity. The time until failure for each component follows an exponential distribution with mean 10. The times until failure for the five components are independent.

What is the expected value of the time that the generator produces electricity.

InteGrand · Sep 3, 2016

leehuan said:
Please explain where their product came from?

Company NewGen has one electirc generator that has five components. All five components are required to work to produce electricity. The time until failure for each component follows an exponential distribution with mean 10. The times until failure for the five components are independent.

What is the expected value of the time that the generator produces electricity.

$\noindent We have $\mathbb{P}\left(X_1 > y \text{ and } X_2 > y \text{ and } \ldots \text{ and } X_5 > y \right) = \mathbb{P} \left(\bigcap _{i = 1}^{5} X_i > y\right) = \prod _{i=1}^{5} \mathbb{P}\left(X_i > y\right)$, since the $X_i$'s are \textbf{independent}.$

$\noindent (Remember the intersection basically means `and', and the probability of intersections of events like this for independent r.v.'s is the product of the individual events.)$

$\noindent Btw, using the same method, we can show that if $X_1,\ldots X_n$ are independent exponential r.v.'s with respective rate parameters $\lambda_1, \ldots ,\lambda_n$ (where $n$ is a given positive integer), and $Y = \min _{i = 1,\ldots n} X_i$, then $Y$ is exponential with rate parameter $\lambda$, where $\lambda = \sum_{i = 1}^{n} \lambda_i$. (What the working in that image has shown is that $Y\sim \mathrm{Exponential} (5\lambda)$.)$

$\noindent Also, I think there's a slight typo in the solution towards the end. They left out a $y$ in exponents. The term in the product should be $e^{-\lambda y}$ and the final expression should be $e^{-5\lambda y}$.$

ACTL1101 Questions Help (mostly first year uni probability) (1 Viewer)

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