My Math Forum Deducing probability given P(x=a)=P(x=b) and Var(x)=1?

February 5th, 2017, 04:27 PM   #1
Newbie

Joined: Aug 2013

Posts: 18
Thanks: 0

Deducing probability given P(x=a)=P(x=b) and Var(x)=1?

Hi, would appreciate some help on these probability problems. For question 1, wouldn't the probability of P(X=0)=1, since it's just the same as the latter two given probabilities? Sorry, could someone correct my logic as I'm sure I am completely off. I'm not sure how to use the Variance for this problem.

For question 2, I integrated f(x) from 0 to w, and equated it to 1, to get c=6/(3w^3-2w^4).
Then, to get E[Y], I plugged c back into the function and integrated it again, with an extra x in the equation to get the value (3w^2-4w)/(2w-3).
Finally, Var[10Y+1]=100Var(Y)=100[E[x^2]-(E[x])^2], which at this point, seems like a monster to compute, so I am doubting my answers.

Questions 3 and 5 are not exactly in the same category, but I would appreciate some pointers if anyone could help.

Question 4, is this just a case of using mean=1, standard deviation=2, and since X=Normal(1,2), is E[X^3] just the mean=1 cubed, which is still 1? Sorry, I know there's a lot missing here.

Would appreciate any help!
Attached Images
 IMG_0791.jpg (83.3 KB, 9 views)

 February 5th, 2017, 07:51 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 1,944 Thanks: 1011 (1) $P[2]=P[-2]=a$ $2a + P[0]=1$ $P[0]=1-2a$ $Var[X] = E[X^2] - E[X]^2 =$ $E[X^2] = 2^2a + (-2)^2 a + 0 = 8a$ $E[X] = 2a -2a + 0 = 0$ $Var[X] = 8a =1$ $a = \dfrac 1 8$ $P[0] = 1 - 2a = 1 - \dfrac{2}{8} = \dfrac 3 4$ Thanks from facebook Last edited by romsek; February 5th, 2017 at 08:13 PM.
 February 5th, 2017, 08:04 PM #3 Senior Member     Joined: Sep 2015 From: USA Posts: 1,944 Thanks: 1011 (2) As you did $\displaystyle{\int_0^2}~c x(w-x)~dx = 1 \Rightarrow c = \dfrac{6}{w^3}$ looks like you had an calculus error somewhere. $E[Y] = \displaystyle{\int_0^w}~y~ c y(w-y)~dy=\dfrac {6}{w^3} \displaystyle{\int_0^w}~y^2(w-y)~dy = \dfrac w 2$ $Var[10Y+1] = (10^2)Var[Y] + Var[1] = 10^2 Var[Y]$ $Var[Y] = E[Y^2]-E[Y]^2 = \displaystyle{\int_0^w}~y^2 f_Y(y)~dy - \left(\dfrac w 2 \right)^2 = \dfrac{w^2}{20}$ I leave it to you to finish and to supply all the integration details. Thanks from facebook
 February 5th, 2017, 08:12 PM #4 Senior Member     Joined: Sep 2015 From: USA Posts: 1,944 Thanks: 1011 (4) $X \sim N(1,2)$ Probably easiest to view $X$ as $X = 2Y+1$ where $Y \sim N(0,1)$ $E[X^3] = E[(2Y+1)^3] = E[8 Y^3+12 Y^2+6 Y+1] =$ $8E[Y^3] + 12 E[Y^2] + 6 E[Y] + 1$ Odd moments of the Standard Normal distribution are 0 as the distribution is symmetric about 0. So this becomes $E[Y^3] = 12 E[Y^2] + 1 = 12(1^2) + 1 = 13$ Thanks from facebook
 February 5th, 2017, 09:38 PM #5 Newbie   Joined: Aug 2013 Posts: 18 Thanks: 0 Ah yes, I did have a calculation error along the way. You have been my saving grace. Thank you so much for your help!
February 5th, 2017, 10:21 PM   #6
Senior Member

Joined: Sep 2015
From: USA

Posts: 1,944
Thanks: 1011

Quote:
 Originally Posted by romsek $\displaystyle{\int_0^2}~c x(w-x)~dx = 1 \Rightarrow c = \dfrac{6}{w^3}$
should be

$\displaystyle{\int_0^w}~c x(w-x)~dx = 1 \Rightarrow c = \dfrac{6}{w^3}$

 Tags deducing, probability, pxapxb, varx1

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post facebook Probability and Statistics 1 February 5th, 2017 08:33 PM rivaaa Advanced Statistics 5 November 5th, 2015 01:51 PM hbonstrom Applied Math 0 November 17th, 2012 07:11 PM token22 Advanced Statistics 2 April 26th, 2012 03:28 PM naspek Calculus 1 December 15th, 2009 01:18 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top