
#1
Feb2112, 09:24 AM

P: 29

Let Y ~ Bi(m,q).
Use the definition of E[g(Y)] to derive E[Y^2]. Hint: Write Y^2 as Y(Y1) + Y. You do not have to rederive E[Y]. Not sure where to start with this; my initial reaction was to use the moment generating function? 



#2
Feb2112, 10:45 AM

Mentor
P: 16,690

Please post such questions in the homework forum. I moved it for you.




#3
Feb2112, 10:49 AM

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RGV 



#4
Feb2112, 01:46 PM

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Use the definition of E[g(Y)] to derive E[Y^2] 



#5
Feb2112, 02:53 PM

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RGV 



#6
Feb2112, 07:29 PM

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P: 3,175

The meaning of the direction to "use the definition of E(g(y))" would be clearer if mathmajor23 explained how "E(g(y))" has been used in his course materials. 



#7
Feb2112, 10:12 PM

P: 29

By definition, E[g(Y)] = the summation over all values of y of [g(y)p(y)]




#8
Feb2212, 02:14 AM

Sci Advisor
P: 3,175

I speculate that the problem wants you to show your virtuosity with sums and binomial coefficients. (That's my attempt to mindread what its author wants. If this problem is from the chapter on moment generating functions, it might want you to use them or it might want you do things "the hard way" so you'll appreciate moment generating functions!)
If we begin from the definition: [itex] E(Y^2) = \sum_{k = 0}^m \ k^2 \ \binom{m}{k} q^k (1q)^{mk} [/itex] and use the hint: [itex] = \sum_{k=0}^m \ (k(k1) + k) \ \binom{m}{k} q^k (1q)^{mk} [/itex] [itex] = \sum_{k=0}^m \ k(k1) \binom{m}{k} q^k (1q)^{mk} + \sum_{k=0}^m \ k \binom{m}{k} q^k (1q)^{mk} [/itex] [itex] = \sum_{k=0}^m \ k(k1) \ \frac{m!}{(mk)! k!} q^k (1q)^{mk} + mq [/itex] Does canceling the factors of [itex] k [/itex] let us rewrite the factors [itex] k(k1) \binom{m}{k} q^k (1q)^{mk} [/itex] as [itex] q (1q) m(k1) \binom{m1}{k1} q^{k1} (1q)^{m1k}[/itex] ? If, so the summation begins to look like a linear expression in the expectation of a binomial with parmeters [itex] m1, q [/itex]. (Edit, This morning, I think that we should only factor out one of the factors [itex] q [/itex] and [itex](1q) [/itex] to get a Bi[itex](m1,q) [/itex].) It probably won't be that simple since you can't "cancel" out [itex] k [/itex] when [itex] k = 0 [/itex], but since the term is 0 you can leave it out of the summation. We'll have to pay attention to what the indexes of summation are. We might get an expression that involves the mean of bi(m1,q) minus a few terms. I began looking at this problem before it got moved to the homework section. I suppose now there's an excuse to leave it to you to see if that idea works. 



#9
Feb2212, 11:22 AM

P: 29

You're right! My professor "loves" MGF and says she will make us appreciate them, and this is like the third proof we have to do with these nasty summations. I follow you all the way up to the point where you try canceling the factors of k, that's where I am getting confused. I will try rewriting the factors as you recommend and see where I go from there.




#10
Feb2212, 12:30 PM

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What methods are you permitted to use? If I were not allowed to use an MGF I would, instead, assume a form like EN^2 = a*n^2 + b*n + c, then find a, b, c by looking at explicit values of EN^2 for small values of n, such as n = 1, 2, 3. Then I would try to _prove_ correctness of the result, by induction, for example. As regards induction: if we let b[n,k] = C(n,k)*p^k (1p)^(nk), then we have k*b[n,k] = n*p* b[n1,k1], etc.
RGV 



#11
Feb2212, 10:32 PM

P: 29

My professor wants us to use just definitions and summations.
As to the last step of the ∑mk=0 k(k−1) m!(m−k)!k!qk(1−q)m−k+mq , the next thing to do is to cancel out the k's Edit: I can't seem to get the summation symbol to agree with me on here :p 



#12
Feb2212, 10:40 PM

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RGV 



#13
Feb2412, 11:18 PM

P: 26

You can derive E(Y^2) easily:
let f(x) = ƩC(m,x)*q^x * (1q)^(mx) E(X^2) = ƩC(m,x)*x^2 *q^x * (1q)^(mx) Observe that this is equivalent to ƩC(m1,x1)*(m/x)*q^x*(1q)^(mx)*x^2 This becomes: ƩC(m1,x1)*(m)*q^x*(1q)^(mx)*x Let y = x1 x = y+1 ƩC(m1,y)*m*q^(y+1)*(1q)^(m1y)*(y+1) This equals ƩC(m1,y)*m*q*q^(y)*(1q)^(m1y)*(y+1) This equals mq*ƩC(m1,y)*q^(y)*(1q)^(m1y)*(y+1) Expand via factor (y+1) mq*ƩC(m1,y)*q^(y)*(1q)^(m1y)*(y) + mq*ƩC(m1,y)*q^(y)*(1q)^(m1y)*(1) The term on the right is simply mq, because the sum itself is equal to 1 The term on the left can be solved easily: mq*ƩC(m1,y)*q^(y)*(1q)^(m1y)*(y) becomes.. mq*ƩC(m2,y1)*((m1)/y)*q^(y)*(1q)^(m1y)*(y) =mq*ƩC(m2,y1)*(m1)*q^(y)*(1q)^(m1y) =(m1)*mq*ƩC(m2,y1)*q^(y)*(1q)^(m1y) Let z = y1 z+1=y (m1)*mq*ƩC(m2,z)*q^(z+1)*(1q)^(m2z) =(m1)*mq^2*ƩC(m2,z)*q^(z)*(1q)^(m2z) = (m1)*mq^2 So we have.. ƩC(m,x)*x^2 *q^x * (1q)^(mx) = (m1)*mq^2 + mq 



#14
Feb2512, 09:36 AM

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RGV 


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