Statistics Marathon & Questions (5 Viewers)

davidgoes4wce · Apr 21, 2017

Re: University Statistics Discussion Marathon

He-Mann said:
Not sure... how to compute... (p = 0.5)

$X \sim \mathrm{Bin}(78, p) \implies P(X \geq 43) = \sum_{k=43}^{78} \binom{78}{k}p^k (1-p)^{78-k}$

what value of p did you get?

InteGrand · Apr 21, 2017

Re: University Statistics Discussion Marathon

davidgoes4wce said:
what value of p did you get?

Do you mean the value of the sum? (The "p" He-Mann used was referring to the success probability parameter, which is 0.5.)

leehuan · Apr 21, 2017

Re: Statistics

U₁, ..., U_n are iid Unif(0,1)

$Y_n := n \min\{U_1,\dots,U_n\}\\ \text{RTP: }F_{Y_n}(y) = 1-\left(1-\frac{y}{n}\right)^n$

So I followed their working up to

$F_{Y_n}(y) = \dots = \mathbb{P}\left(U_1\le \frac{y}{n}, \dots, U_n \le \frac{y}{n}\right)$

$\text{But then for some reason the solutions said that this was equal to }\\ 1-\mathbb{P}\left(U_1>\frac{y}{n}, \dots, U_n > \frac{y}{n}\right) = 1-\left(\int_{\frac{y}n}^1 1\, du\right)^n\\ \text{to get the answer (I understand how they used independence)}$

$\text{What I want to know is why I can't say that this is just equal to }\left(\int_0^\frac{y}{n} 1\, du\right)^n\\ \text{Because that gets me just }\left(\frac{y}{n}\right)^n\text{ which is wrong}$
______________________

Edit: Actually my bad on transcription error, their working stopped here

$\mathbb{P}\left(\min \{U_1,\dots,U_n)\} \le \frac{y}{n}\right)$

InteGrand · Apr 21, 2017

Re: Statistics

leehuan said:
U₁, ..., U_n are iid Unif(0,1)

$Y_n := n \min\{U_1,\dots,U_n\}\\ \text{RTP: }F_{Y_n}(y) = 1-\left(1-\frac{y}{n}\right)^n$

So I followed their working up to

$F_{Y_n}(y) = \dots = \mathbb{P}\left(U_1\le \frac{y}{n}, \dots, U_n \le \frac{y}{n}\right)$

$\text{But then for some reason the solutions said that this was equal to }\\ 1-\mathbb{P}\left(U_1>\frac{y}{n}, \dots, U_n > \frac{y}{n}\right) = 1-\left(\int_{\frac{y}n}^1 1\, du\right)^n\\ \text{to get the answer (I understand how they used independence)}$

$\text{What I want to know is why I can't say that this is just equal to }\left(\int_0^\frac{y}{n} 1\, du\right)^n\\ \text{Because that gets me just }\left(\frac{y}{n}\right)^n\text{ which is wrong}$
______________________

Edit: Actually my bad on transcription error, their working stopped here

$\mathbb{P}\left(\min \{U_1,\dots,U_n)\} \le \frac{y}{n}\right)$

The integral you were thinking about is actually the "complementary CDF" or "survival function" (though the limits aren't right and I realised I misinterpreted your query before seeing your addendum), which is the probability Y_n > y. It's actually neater here find this survival function first and then just get the CDF of Y_n from that by taking its complement (subtracting it from 1). The reason it's nicer to work with the survival function here is that's we're dealing with a min function, and the minimum of some numbers is greater than something iff all the numbers are greater than that thing. If instead we were working with a max function, it'd be nicer to use the CDF, because a maximum of some numbers is less than or equal to something iff all the numbers are less than or equal to that thing.

Edit: I think I misinterpreted your query (I think I get it now after seeing your addendum). I addressed what I think was your query in my next post.

InteGrand · Apr 21, 2017

Re: Statistics

leehuan said:
U₁, ..., U_n are iid Unif(0,1)

$Y_n := n \min\{U_1,\dots,U_n\}\\ \text{RTP: }F_{Y_n}(y) = 1-\left(1-\frac{y}{n}\right)^n$

So I followed their working up to

$F_{Y_n}(y) = \dots = \mathbb{P}\left(U_1\le \frac{y}{n}, \dots, U_n \le \frac{y}{n}\right)$

$\text{But then for some reason the solutions said that this was equal to }\\ 1-\mathbb{P}\left(U_1>\frac{y}{n}, \dots, U_n > \frac{y}{n}\right) = 1-\left(\int_{\frac{y}n}^1 1\, du\right)^n\\ \text{to get the answer (I understand how they used independence)}$

$\text{What I want to know is why I can't say that this is just equal to }\left(\int_0^\frac{y}{n} 1\, du\right)^n\\ \text{Because that gets me just }\left(\frac{y}{n}\right)^n\text{ which is wrong}$
______________________

Edit: Actually my bad on transcription error, their working stopped here

$\mathbb{P}\left(\min \{U_1,\dots,U_n)\} \le \frac{y}{n}\right)$

$\noindent I'm not 100\% sure what you're asking (possibly due to your transcription error). But I think you were asking why you can't just say $\mathbb{P}\left(\min \left\{U_1,\ldots,U_n\right\} \leq \frac{y}{n}\right) = \left(\mathbb{P}\left(U \leq \frac{y}{n}\right)\right)^{n}$. For this to be true, we'd want it to be true that $\min \left\{U_1,\ldots,U_n\right\} \leq \frac{y}{n}$ iff all $U_{i}\leq \frac{y}{n}$, but this is \textbf{not} the case with minimums and $\leq$ signs. Instead, work with $>$ than signs (i.e. survival functions), because it \emph{is} true of minimums that $\min_{1\leq i \leq n}a_i > t$ iff all $a_{i} > t$.$

leehuan · Apr 21, 2017

Re: Statistics

InteGrand said:
$\noindent I'm not 100\% sure what you're asking (possibly due to your transcription error). But I think you were asking why you can't just say $\mathbb{P}\left(\min \left\{U_1,\ldots,U_n)\right\} \leq \frac{y}{n}\right) = \left(\mathbb{P}\left(U \leq \frac{y}{n}\right)\right)^{n}$. For this to be true, we'd want it to be true that $\min \left\{U_1,\ldots,U_n)\right\} \leq \frac{y}{n}$ iff all $U_{i}\leq \frac{y}{n}$, but this is \textbf{not} the case with minimums and $\leq$ signs. Instead, work with $>$ than signs (i.e. survival functions), because it \emph{is} true of minimums that $\min_{1\leq i \leq n}a_i > t$ iff all $a_{i} > t$.$

Got it, thanks, 90% sure that was my confusion

leehuan · Apr 21, 2017

Re: Statistics

I computed an MGF to be

$m_X(u) = \frac{2}{u\theta^2} \left(\theta e^{u\theta}+\frac{1-e^{u\theta}}u\right)$

Therefore it does not exist because it's not defined for any u in [interval containing 0] right?

InteGrand · Apr 21, 2017

Re: Statistics

leehuan said:
I computed an MGF to be

$m_X(u) = \frac{2}{u\theta^2} \left(\theta e^{u\theta}+\frac{1-e^{u\theta}}u\right)$

Therefore it does not exist because it's not defined for any u in [interval containing 0] right?

Actually that function has finite limit as u -> 0, as you can show using L'Hôpital's rule for example. So the MGF does exist.

leehuan · Apr 21, 2017

Re: Statistics

InteGrand said:
Actually that function has finite limit as u -> 0, as you can show using L'Hôpital's rule for example. So the MGF does exist.

Nice. I knew I was missing something.

He-Mann · Apr 21, 2017

Re: University Statistics Discussion Marathon

Can't any quick way of doing it at the moment. Please let me know if any of this is wrong.

$P(X \geq 43) = \sum_{k = 43}^{78} \binom{78}{k} p^k (1-p)^{78 -k} = 1 - \sum_{k=0}^{42} \binom{78}{k} p^k (1-p)^{78-k}$

Since probabilities sum to 1, we have

$\left(\sum_{k=0}^{38}\binom{78}{k} p^k (1-p)^{78-k} \right) + \left(\binom{78}{39} p^{39} (1-p)^{39} \right) + \left(\sum_{k=40}^{78} \binom{78}{k}p^k (1-p)^{78-k} \right) = 1 \quad \dots (*)$

By symmetry (i.e. Pascal's Triangle), we have

$\sum_{k=0}^{38}\binom{78}{k} p^k (1-p)^{78-k} = \sum_{k=40}^{78} \binom{78}{k}p^k (1-p)^{78-k}$

Then from (*), we have

$\begin{align*}\sum_{k=0}^{38}\binom{78}{k} p^k (1-p)^{78-k} &= \frac{1}{2}\binom{78}{39} p^{39} (1-p)^{39} \\&= \frac{1}{2} \binom{78}{39}(0.5)^{78} \approx 0.045 \\\\ \textbf{MISTAKE HAPPENS ABOVE. THE RHS SHOULD BE }&=\frac{1}{2} (1 - \binom{78}{39}(0.5)^{78}) \\ &\approx 0.95 \end{align*}$

Now, we compute the desired probability,

$\begin{align*}P(X \geq 43)& = 1 - \sum_{k=0}^{42} \binom{78}{k} p^k (1-p)^{78-k} \\ & \approx 1 - \left( 0.045 + 2\times 0.045+ \sum_{k=40}^{42}\binom{78}{k} p^k (1-p)^{78-k}\right) \\ & = 1 - \left( 0.045 + 2\times 0.045 + 0.2409 \right) \\& = 0.6241 \end{align*}$

In retrospect, I should have wrote p = 0.5 = 1 - p and combined the product which would make it easier to read.

leehuan · Apr 22, 2017

Re: Statistics

$f_{V,W}(v,w) = \frac{1}{3600}, \, 0 \le v,w < 60$

$\text{Required: The density of }Z=W-V$

$\text{Approach so far: Additionally, define }U=W\\ \text{After some computation, }f_{Z,U}(z,u) = \frac{1}{3600}\text{ and }0\le u < 60$

Once again stuck on domains. I keep forgetting how linear transforms map regions so maybe just another 'by inspection' method please?

(Intuitive guess: -60 < z < 60. Just don't know how to get there)

InteGrand · Apr 22, 2017

Re: Statistics

leehuan said:
$f_{V,W}(v,w) = \frac{1}{3600}, \, 0 \le v,w < 60$

$\text{Required: The density of }Z=W-V$

$\text{Approach so far: Additionally, define }U=W\\ \text{After some computation, }f_{Z,U}(z,u) = \frac{1}{3600}\text{ and }0\le u < 60$

Once again stuck on domains. I keep forgetting how linear transforms map regions so maybe just another 'by inspection' method please?

(Intuitive guess: -60 < z < 60. Just don't know how to get there)

It's quite easy with the linear transforms method. The sample space in the v-w space is just a square. A square gets mapped to a parallelogram in the u-z space (in general) by a linear transformation like the one you have. (Recall that in general, an invertible linear map from Rⁿ to Rⁿ will map lines to lines and parallelograms to parallelograms.)

To find the coordinates of the vertices this parallelogram, just find what points each of the vertices of the square in the v-w sample space get mapped to by the linear transformation. Then the interior of the sample space in the v-w space gets mapped to the inside of the parallelogram with those vertices in the u-z space.

InteGrand · Apr 22, 2017

Re: Statistics

leehuan said:
$f_{V,W}(v,w) = \frac{1}{3600}, \, 0 \le v,w < 60$

$\text{Required: The density of }Z=W-V$

$\text{Approach so far: Additionally, define }U=W\\ \text{After some computation, }f_{Z,U}(z,u) = \frac{1}{3600}\text{ and }0\le u < 60$

Once again stuck on domains. I keep forgetting how linear transforms map regions so maybe just another 'by inspection' method please?

(Intuitive guess: -60 < z < 60. Just don't know how to get there)

$\noindent Here's a ``by inspection'' method. We know $U = W$, so the range for $U$ will be same as that for $W$, i.e. $0 \leq u < 60$. For $Z$, we know $Z$ is just $W - V \equiv U - V$, so due to the range of values for $V$ and $U$, $Z$ can take values from $U - 60$ up to $U$. So we also will get $u - 60 < z \leq u$. So the sample space in $u$-$z$ space is $\left\{(u,z) \in \mathbb{R}^{2} : 0 \leq u < 60 \text{ and } u-60 < z \leq u\right\}$. This is a parallelogram region bounded by the parallel lines $z = u$ and $z = u - 60$ and the vertical parallel lines $u = 0$ and $u = 60$.$

leehuan · Apr 22, 2017

Re: Statistics

That's interesting, u - 60 < z \le u was the first thing I had.

$\text{But in that case, after I did }f_Z(z) =\int_0^{60} \frac1{3600}du\\ \text{What becomes the domain of }z\text{ then?}$

InteGrand · Apr 22, 2017

Re: Statistics

leehuan said:
That's interesting, u - 60 < z \le u was the first thing I had.

$\text{But in that case, after I did }f_Z(z) =\int_0^{60} \frac1{3600}du\\ \text{What becomes the domain of }z\text{ then?}$

$\noindent Those bounds on the integral aren't correct. For the bounds, you will need to take cases based on whether $z \geq 0$ or $z < 0$ (refer to your diagram of the $u$-$z$ sample space). And in the end, the range of values of $Z$ will be $(-60,60)$. We could see this from the start, essentially since $Z$ is the difference between two joint random variables defined on $\left[0, 60)^{2}$, so their difference is going to be between $-60$ and $+60$.$

leehuan · Apr 22, 2017

Re: Statistics

InteGrand said:
$\noindent Those bounds on the integral aren't correct. For the bounds, you will need to take cases based on whether $z \geq 0$ or $z < 0$ (refer to your diagram of the $u$-$z$ sample space). And in the end, the range of values of $Z$ will be $(-60,60)$. We could see this from the start, essentially since $Z$ is the difference between two joint random variables defined on $\left[0, 60)^{2}$, so their difference is going to be between $-60$ and $+60$.$

Awesome, I finally see it.

InteGrand · Apr 22, 2017

Re: Statistics

leehuan said:
Awesome, I finally see it.

Nice. Incidentally, here's what the graph of the pdf should look like: http://www.graphsketch.com/?eqn1_co..._lines=1&line_width=4&image_w=850&image_h=525 .

Basically a symmetric triangle shaped graph on (-60, 60) that has value 0 at the endpoints (this is enough to specify the height of the triangle too (and hence the pdf), since the total area under the curve must be 1). You may like to think about coming up with an intuitive explanation of why we could have expected this.

leehuan · Apr 22, 2017

Re: Statistics

InteGrand said:
Nice. Incidentally, here's what the graph of the pdf should look like: http://www.graphsketch.com/?eqn1_co..._lines=1&line_width=4&image_w=850&image_h=525 .

Basically a symmetric triangle shaped graph on (-60, 60) that has value 0 at the endpoints (this is enough to specify the height of the triangle too (and hence the pdf), since the total area under the curve must be 1). You may like to think about coming up with an intuitive explanation of why we could have expected this.

Hmm.

The original question defined V and W as the time taken for two different persons have to wait for their train to ride (given that they do not have to wait any more than 1 hour). The fact that V and W were Unif(0,60) could suggest why it's a triangle (and not say the bell curve), and it "decreases in both directions from 0" because

Bad description because my brain seems to not be working, but there would be less 45 min time ranges than say, 20 min time ranges within a 1 hour time span?

He-Mann · Apr 27, 2017

Re: University Statistics Discussion Marathon

Bump! Really interested if someone has thought up of a more efficient way to do the question (or identify, if it exists, the flaws in my work).

Paradoxica · Apr 27, 2017

Re: University Statistics Discussion Marathon

Mathematica 11 gives me 0.2141 from your initial sum...

Statistics Marathon & Questions (5 Viewers)

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