# Thread: Statistics Marathon & Questions

1. ## Re: University Statistics Discussion Marathon

Just want to confirm if this is the right answer :

2. ## Re: University Statistics Discussion Marathon

Originally Posted by davidgoes4wce
Just want to confirm if this is the right answer :

$\noindent I haven't checked your numerical calculations but yes that's the right approach, i.e. standardise the random variable using Z = \frac{X -\mu}{\sigma} so that Z becomes standard normal, correspondingly change the inequality bounds, and then use standard normal CDF values to find the required probabilities (e.g. in a table of values).$

3. ## Re: Statistics

How do you find the marginal distribution of a random variable, given a table of the joint probabilities of that random variable and another?

i.e. fY(y).

I see the solution but not sure where they are getting it from.

4. ## Re: Statistics

Originally Posted by Flop21
How do you find the marginal distribution of a random variable, given a table of the joint probabilities of that random variable and another?

i.e. fY(y).

I see the solution but not sure where they are getting it from.
You integrate or sum the joint density function over all values of the other variable(s).

5. ## Re: Statistics

Originally Posted by Flop21
How do you find the marginal distribution of a random variable, given a table of the joint probabilities of that random variable and another?

i.e. fY(y).

I see the solution but not sure where they are getting it from.
$\text{This formula you're just gonna have to know. Because in our courses we only consider (at most) bivariate distributions}\\ \text{I'll just assume that the other r.v. is }X$

$f_Y(y) = \int_{\mathbb{R}} f_{X,Y}(x,y)\,dx\text{ for continuous}$

$\mathbb{P}(Y=y) = \sum_{\forall x}\mathbb{P}(X=x,Y=y)\text{ for discrete}$

$\text{with use of appropriate boundaries/indicator functions/whatever}$

Which is nothing different to what IG said. He basically just described a general case where you have n r.v.s

6. ## Re: Statistics

$X_1, X_2 \sim \text{exp}(2)\text{ are iid.}$

And we're using the convention X ~ exp(a) => fX(x) = 1/a exp(-x/a)

$\text{Determined in part a) }f_{X_1,X_2}(x_1,x_2)=\frac{1}{4}e^{-\frac{x_1+x_2}2}\text{ for }x_1>0, x_2>0$

$\text{Define }Y_1 := \frac12 (X_1-X_2)\text{ and }Y_2:=X_2\\ \text{Found in part b) }f_{Y_1,Y_2}(y_1,y_2) = \frac12 e^{-(y_1+y_2)}$

But how do you find the conditions on y1 and y2 where this density is defined for?

7. ## Re: University Statistics Discussion Marathon

50 % of the meals sold at Graylands Hospital contain rice.

Answer the following question, keeping in mind that your answer must be a number between 0 and 1, i.e do not use percentages. Round off the final result to 2dp, but don't round numbers during your calculations.

What is the probability that out of 78 meals sold at Graylands Hospital, 43 or more contain rice?

8. ## Re: Statistics

Originally Posted by leehuan
$X_1, X_2 \sim \text{exp}(2)\text{ are iid.}$

And we're using the convention X ~ exp(a) => fX(x) = 1/a exp(-x/a)

$\text{Determined in part a) }f_{X_1,X_2}(x_1,x_2)=\frac{1}{4}e^{-\frac{x_1+x_2}2}\text{ for }x_1>0, x_2>0$

$\text{Define }Y_1 := \frac12 (X_1-X_2)\text{ and }Y_2:=X_2\\ \text{Found in part b) }f_{Y_1,Y_2}(y_1,y_2) = \frac12 e^{-(y_1+y_2)}$

But how do you find the conditions on y1 and y2 where this density is defined for?
$\noindent We just need to find what the region for the original sample space, which is \Omega := \left\{ (x_{1}, x_{2}) \in \mathbb{R}^{2} : x_{1} > 0, x_{2} > 0\right\} (i.e. first quadrant), gets mapped to under the transformation in question, which is \begin{bmatrix}y_{1} \\ y_{2}\end{bmatrix} = \begin{bmatrix}\frac{1}{2}x_{1} -\frac{1}{2}x_{2} \\ x_{2}\end{bmatrix}. Since this is a linear transformation, this is quite easy to do, if you recall basic properties of how linear maps transform regions.$

9. ## Re: Statistics

$\noindent If you want a by inspection'' method, you can observe the new region will be y_{2} > 0 and y_{1} > -\frac{1}{2}y_{2}. This is essentially because Y_{2} = X_{2}, so the range of values for y_{2} is just the same as x_{2}, so is y_{2} > 0. For Y_{1}, we note that Y_{1} = -\frac{1}{2}X_{2} + \frac{1}{2} X_{1} = -\frac{1}{2}Y_{2} + \frac{1}{2}X_{1}. Now, \frac{1}{2}X_{1} has range (0,\infty), so the values Y_{1} can take go from than that of -\frac{1}{2}Y_2 up to (and not including) +\infty. This explains why the other thing was y_{1} > -\frac{1}{2}y_{2}.$

10. ## Re: University Statistics Discussion Marathon

Originally Posted by davidgoes4wce
50 % of the meals sold at Graylands Hospital contain rice.

Answer the following question, keeping in mind that your answer must be a number between 0 and 1, i.e do not use percentages. Round off the final result to 2dp, but don't round numbers during your calculations.

What is the probability that out of 78 meals sold at Graylands Hospital, 43 or more contain rice?

Not sure... how to compute... (p = 0.5)

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

11. ## Re: University Statistics Discussion Marathon

Originally Posted by He-Mann
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?

12. ## Re: University Statistics Discussion Marathon

Originally Posted by davidgoes4wce
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.)

13. ## Re: Statistics

U1, ..., Un 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

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

14. ## Re: Statistics

Originally Posted by leehuan
U1, ..., Un 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.

15. ## Re: Statistics

Originally Posted by leehuan
U1, ..., Un 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

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

16. ## Re: Statistics

Originally Posted by InteGrand
$\noindent I'm not 100\% sure what you're asking (possibly due to your transcription error). But I think you were asking why you can't just say \mathbb{P}\left(\min \left\{U_1,\ldots,U_n)\right\} \leq \frac{y}{n}\right) = \left(\mathbb{P}\left(U \leq \frac{y}{n}\right)\right)^{n}. For this to be true, we'd want it to be true that \min \left\{U_1,\ldots,U_n)\right\} \leq \frac{y}{n} iff all U_{i}\leq \frac{y}{n}, but this is \textbf{not} the case with minimums and \leq signs. Instead, work with > than signs (i.e. survival functions), because it \emph{is} true of minimums that \min_{1\leq i \leq n}a_i > t iff all a_{i} > t.$
Got it, thanks, 90% sure that was my confusion

17. ## 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?

18. ## Re: Statistics

Originally Posted by leehuan
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.

19. ## Re: Statistics

Originally Posted by InteGrand
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.

20. ## 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.

21. ## 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)

22. ## Re: Statistics

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

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

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

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.

23. ## Re: Statistics

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

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

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

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

24. ## Re: Statistics

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

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

25. ## Re: Statistics

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

$\text{But in that case, after I did }f_Z(z) =\int_0^{60} \frac1{3600}du\\ \text{What becomes the domain of }z\text{ then?}$
$\noindent Those bounds on the integral aren't correct. For the bounds, you will need to take cases based on whether z \geq 0 or z < 0 (refer to your diagram of the u-z sample space). And in the end, the range of values of Z will be (-60,60). We could see this from the start, essentially since Z is the difference between two joint random variables defined on \left[0, 60)^{2}, so their difference is going to be between -60 and +60.$

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