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February 5th, 2017, 04:27 PM   #1
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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!
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February 5th, 2017, 07:51 PM   #2
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(1)

$P[2]=P[-2]=a$

$2a + P[0]=1$

$P[0]=1-2a$

$Var[X] = E[X^2] - E[X]^2 =$

$E[X^2] = 2^2a + (-2)^2 a + 0 = 8a$

$E[X] = 2a -2a + 0 = 0$

$Var[X] = 8a =1$

$a = \dfrac 1 8$

$P[0] = 1 - 2a = 1 - \dfrac{2}{8} = \dfrac 3 4$
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February 5th, 2017, 08:04 PM   #3
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(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.
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February 5th, 2017, 08:12 PM   #4
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(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

$E[Y^3] = 12 E[Y^2] + 1 = 12(1^2) + 1 = 13$
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February 5th, 2017, 09:38 PM   #5
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Ah yes, I did have a calculation error along the way. You have been my saving grace. Thank you so much for your help!
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February 5th, 2017, 10:21 PM   #6
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Quote:
Originally Posted by romsek View Post
$\displaystyle{\int_0^2}~c x(w-x)~dx = 1 \Rightarrow c = \dfrac{6}{w^3}$
should be

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