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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^32w^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^24w)/(2w3). 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! 
February 5th, 2017, 07:51 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 1,944 Thanks: 1011 
(1) $P[2]=P[2]=a$ $2a + P[0]=1$ $P[0]=12a$ $Var[X] = E[X^2]  E[X]^2 =$ $E[X^2] = 2^2a + (2)^2 a + 0 = 8a$ $E[X] = 2a 2a + 0 = 0$ $Var[X] = 8a =1$ $a = \dfrac 1 8$ $P[0] = 1  2a = 1  \dfrac{2}{8} = \dfrac 3 4$ 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 $\displaystyle{\int_0^2}~c x(wx)~dx = 1 \Rightarrow c = \dfrac{6}{w^3}$ looks like you had an calculus error somewhere. $E[Y] = \displaystyle{\int_0^w}~y~ c y(wy)~dy=\dfrac {6}{w^3} \displaystyle{\int_0^w}~y^2(wy)~dy = \dfrac w 2$ $Var[10Y+1] = (10^2)Var[Y] + Var[1] = 10^2 Var[Y]$ $Var[Y] = E[Y^2]E[Y]^2 = \displaystyle{\int_0^w}~y^2 f_Y(y)~dy  \left(\dfrac w 2 \right)^2 = \dfrac{w^2}{20}$ I leave it to you to finish and to supply all the integration details. 
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 $E[Y^3] = 12 E[Y^2] + 1 = 12(1^2) + 1 = 13$ 
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  

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deducing, probability, pxapxb, varx1 
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