Statistics Marathon & Questions (2 Viewers)

InteGrand · Apr 27, 2017

Re: University Statistics Discussion Marathon

He-Mann said:
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

$\begin{align*}\sum_{k=0}^{38}\binom{78}{k} p^k (1-p)^{78-k} &= \frac{1}{2}\binom{78}{39} p^{39} (1-p)^{39} \\&= \frac{1}{2} \binom{78}{39}(0.5)^{78} \approx 0.045 \end{align*}$

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.

He-Mann said:
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).

Note that the answer to the sum cannot exceed 0.5 (in general for X ~ Bin(n, 1/2), if k > n/2, then Pr(X ≥ k) ≤ 0.5).

He-Mann · Apr 28, 2017

Re: University Statistics Discussion Marathon

I have corrected my original working. Thanks for the checks, I knew something was strange.

Below is an easier-to-read version.
___________________________________________________________________________

Given that p = 0.5 = 1-p,

$P(X \geq 43) = \sum_{k = 43}^{78} \binom{78}{k} p^k (1-p)^{78 -k} = 1 - (0.5)^{78} \sum_{k=0}^{42} \binom{78}{k}$

By Binomial Theorem (with x = 1) and Pascal's Triangle (i.e. C(n,k) = C(n, n-k)),

$\begin{align*}(1+x)^{78} = \sum_{k=0}^{78} \binom{78}{k} = 2^{78} &\iff 2\left(\sum_{k=0}^{38} \binom{78}{k} \right) + \binom{78}{39} = 2^{78} \\& \iff \sum_{k=0}^{38} \binom{78}{k} = \frac{2^{78} - \binom{78}{39}}{2}\end{align*}$

Now, we compute the desired probability,

$\begin{align*}P(X \geq 43)& = 1 - (0.5)^{78} \left( \left[\sum_{k=0}^{38} \binom{78}{k} \right] + \binom{78}{39} + \binom{78}{40} + \binom{78}{41} + \binom{78}{42} \right)\\&\approx 0.2141\end{align*}$

Flop21 · May 29, 2017

Re: Statistics

How do I find E(XY) of discrete random variable? (I'm trying to find co variance).

The variables are NOT independent.

InteGrand · May 29, 2017

Re: Statistics

Flop21 said:
How do I find E(XY) of discrete random variable? (I'm trying to find co variance).

The variables are NOT independent.

$\noindent If $X$ and $Y$ are discrete, then if you know their joint pmf $f(x,y) = \mathbb{P}\left(X = x,Y = y\right)$, we would have $\mathbb{E}\left[XY\right]$ be given as a double sum over all values in the range of $X$ and $Y$, that is,$

$\mathbb{E}\left[XY\right] = \sum_{\text{all }x}\sum_{\text{all }y}xy\,\mathbb{P}\left(X = x,Y = y\right).$

Flop21 · May 29, 2017

Re: Statistics

Thanks!!

Also what is this:

P(X = 0 | X < 1)

What are they called, with the same random variable twice? I can't see this in my textbook and have no clue what keywords to search.

Thanks.

InteGrand · May 29, 2017

Re: Statistics

Flop21 said:
Thanks!!

Also what is this:

P(X = 0 | X < 1)

What are they called, with the same random variable twice? I can't see this in my textbook and have no clue what keywords to search.

Thanks.

That's the conditional probability that X equals 0 given X < 1.

Flop21 · May 29, 2017

Re: Statistics

Sorry also where do I find an explanation of doing these sorts of problems too:

Tabulate the probability function of Y = X^2 + 1, when given fX(x) table.

No clue where to start here.

InteGrand · May 29, 2017

Re: Statistics

Flop21 said:
Sorry also where do I find an explanation of doing these sorts of problems too:

Tabulate the probability function of Y = X^2 + 1, when given fX(x) table.

No clue where to start here.

You would need to find the range of Y based on the range of X. Also, the probability Y takes on some value would be easily to see from the probability distribution of X. For example, if X took the value 1 with probability 1/2 and 0 with probability 1/2, then Y would take the value 0^2 + 1 = 1 with probability 1/2 (because this happens iff X = 0), and Y would take the value 1^2 + 1 = 2 with probability 1/2 (since this happens iff X = 1).

Flop21 · May 30, 2017

Re: Statistics

InteGrand said:
You would need to find the range of Y based on the range of X. Also, the probability Y takes on some value would be easily to see from the probability distribution of X. For example, if X took the value 1 with probability 1/2 and 0 with probability 1/2, then Y would take the value 0^2 + 1 = 1 with probability 1/2 (because this happens iff X = 0), and Y would take the value 1^2 + 1 = 2 with probability 1/2 (since this happens iff X = 1).

Thank you

so for example this question:

I understand what to do, but what do I do when I get two of the same values for Y? E.g. in this I get 2 -> 1/4 and 2 -> 2. The answers just have one column for 2 and it -> 1/8. So do you just multiply the results together to get one entry for Y = 2?

InteGrand · May 30, 2017

Re: Statistics

Flop21 said:
Thank you

so for example this question:

I understand what to do, but what do I do when I get two of the same values for Y? E.g. in this I get 2 -> 1/4 and 2 -> 2. The answers just have one column for 2 and it -> 1/8. So do you just multiply the results together to get one entry for Y = 2?

If there are two values of X that yield the same value for Y, then the probability of Y taking that value is just the sum of X taking on each of the values that gave that Y value.

The probability entry for Y = 2 should be 1/4, because Y = 2 occurs iff X = -1 or X = 1, and

Pr(X = -1 or X = 1) = Pr(X = -1) + Pr(X = 1)

= 1/8 + 1/8

= 1/4.

Flop21 · May 30, 2017

Re: Statistics

InteGrand said:
That's the conditional probability that X equals 0 given X < 1.

Following on with these types of questions..

I'm given fX(x) = 2x, 0 < x < 1.

Calculate P(X < 1/2 | X > 1/3).

So I've realised there's a formula, P(B | A) = P(A and B)/P(A). Which I can use but I get P (X > 1/3) on the bottom. How do I calculate that probability with a ">" with a continuous variable?

Thanks

Flop21 · May 30, 2017

Re: Statistics

Flop21 said:
Following on with these types of questions..

I'm given fX(x) = 2x, 0 < x < 1.

Calculate P(X < 1/2 | X > 1/3).

So I've realised there's a formula, P(B | A) = P(A and B)/P(A). Which I can use but I get P (X > 1/3) on the bottom. How do I calculate that probability with a ">" with a continuous variable?

Thanks

What do I even search to see people doing examples of these questions, searching conditional probability just brings up super simple stuff.

I need examples of people doing like P( X + Y < 30) etc.

InteGrand · May 30, 2017

Re: Statistics

Flop21 said:
Following on with these types of questions..

I'm given fX(x) = 2x, 0 < x < 1.

Calculate P(X < 1/2 | X > 1/3).

So I've realised there's a formula, P(B | A) = P(A and B)/P(A). Which I can use but I get P (X > 1/3) on the bottom. How do I calculate that probability with a ">" with a continuous variable?

Thanks

You can do it using that conditional probability formula you wrote, you'll just need to calculate each of the terms that appear using the given PDF.

InteGrand · May 30, 2017

Re: Statistics

To find Pr(X > 1/3) for that one, we'd just integrate the PDF of X (f(x) = 2x) from x = 1/3 to x = 1.

Flop21 · May 31, 2017

Re: Statistics

Why is the answer (0,4]?

for k = 4, you get 3/4 < x <= 4. So wouldn't 3/4 be the lower limit?

InteGrand · May 31, 2017

Re: Statistics

Flop21 said:
Why is the answer (0,4]?

for k = 4, you get 3/4 < x <= 4. So wouldn't 3/4 be the lower limit?

$\noindent It is not the case that $\bigcap_{k=1}^{\infty}A_{k} = (0,4]$ if $A_{k} = \left(1 - \frac{1}{k}, k\right]$. We want the intersection (all points in common) of all the intervals $\left(1 - \frac{1}{k}, k\right]$ (for $k = 1,2,3,\ldots$).$

$\noindent The point $4$ is in $(0, 4]$ but is not in $A_{1} = \left(0,1]$ and hence not in $\bigcap_{k=1}^{\infty}A_{k}$, so $(0, 4]$ is not the answer.$

Flop21 · May 31, 2017

Re: Statistics

InteGrand said:
$\noindent It is not the case that $\bigcap_{k=1}^{\infty}A_{k} = (0,4]$ if $A_{k} = \left(1 - \frac{1}{k}, k\right]$. We want the intersection (all points in common) of all the intervals $\left(1 - \frac{1}{k}, k\right]$ (for $k = 1,2,3,\ldots$).$

$\noindent The point $4$ is in $(0, 4]$ but is not in $A_{1} = \left(0,1]$ and hence not in $\bigcap_{k=1}^{\infty}A_{k}$, so $(0, 4]$ is not the answer.$

Oops!!

I snapped the wrong question. And posted this in the wrong thread... lol. This is discrete, my bad.

I mean k = 1...4 and it's a union not an intersection.

InteGrand · May 31, 2017

Re: Statistics

Flop21 said:
Oops!!

I snapped the wrong question. And posted this in the wrong thread... lol. This is discrete, my bad.

I mean k = 1...4 and it's a union not an intersection.

$\noindent If it's the union, we basically want the set of all points that belong to at least one of the $A_{k}$. The lower limit would be $0$, but the upper limit is $\infty$. So the union of those $A_{k}$'s is $\left(0,\infty)$.$

Edit: just noticed it was only k = 1 to 4. In that case, the upper limit is 4, so the answer is (0, 4].

Union just means we want the points to be in at least one of the A_k.

He-Mann · May 31, 2017

Re: Statistics

InteGrand said:
To find Pr(X > 1/3) for that one, we'd just integrate the PDF of X (f(x) = 2x) from x = 1/3 to x = 1.

@Flop21, do you understand why we integrate the PDF of X from x = 1/3 to x = 1?

leehuan · Jun 2, 2017

Re: Statistics

Getting lost in what it means to converge in distribution to a constant

$X_n \stackrel{d}{\to} X\text{ is defined to mean}\\ \lim_{n\to \infty}F_{X_n}(x) = F_X(x)$

But I'm not seeing how F_X(x) is allowed to be a constant. Because doesn't F_X(x) technically have to satisfy these?

$F(-\infty) = 0, \quad F(\infty) = 1$

(ignoring the abuse of notation)

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