Computing $ E[Y^2] $ when $Y$ is a piecewise function of $X sim Pois(2) $
$begingroup$
$ X sim Pois(2) $ and let$ Y $ be a random variable : $ Y = begin{cases} 2X , X le 3 \ X , X ge 4 end{cases} $
compute $ E[Y^2] $ ?
I have calculated that $ E[Y] = E[X] + 10e^{-2} $ and I have the identity $ Var(X) = E[X^2] - (E[X])^2 $ and because $ X $ is poisson then $ Var(X) = E[X] $. but I'm not sure if $ Y $ is also a poisson variable?
Any suggestions?
probability poisson-distribution
edited Dec 9 '18 at 16:59
Nate Eldredge
63.6k682171
asked Dec 9 '18 at 16:37
bm1125bm1125
64916
$endgroup$
add a comment |
$begingroup$
$ X sim Pois(2) $ and let$ Y $ be a random variable : $ Y = begin{cases} 2X , X le 3 \ X , X ge 4 end{cases} $
compute $ E[Y^2] $ ?
I have calculated that $ E[Y] = E[X] + 10e^{-2} $ and I have the identity $ Var(X) = E[X^2] - (E[X])^2 $ and because $ X $ is poisson then $ Var(X) = E[X] $. but I'm not sure if $ Y $ is also a poisson variable?
Any suggestions?
probability poisson-distribution
edited Dec 9 '18 at 16:59
Nate Eldredge
63.6k682171
asked Dec 9 '18 at 16:37
bm1125bm1125
64916
$endgroup$
add a comment |
$begingroup$
$ X sim Pois(2) $ and let$ Y $ be a random variable : $ Y = begin{cases} 2X , X le 3 \ X , X ge 4 end{cases} $
compute $ E[Y^2] $ ?
I have calculated that $ E[Y] = E[X] + 10e^{-2} $ and I have the identity $ Var(X) = E[X^2] - (E[X])^2 $ and because $ X $ is poisson then $ Var(X) = E[X] $. but I'm not sure if $ Y $ is also a poisson variable?
Any suggestions?
probability poisson-distribution
edited Dec 9 '18 at 16:59
Nate Eldredge
63.6k682171
asked Dec 9 '18 at 16:37
bm1125bm1125
64916
$endgroup$
$ X sim Pois(2) $ and let$ Y $ be a random variable : $ Y = begin{cases} 2X , X le 3 \ X , X ge 4 end{cases} $
compute $ E[Y^2] $ ?
I have calculated that $ E[Y] = E[X] + 10e^{-2} $ and I have the identity $ Var(X) = E[X^2] - (E[X])^2 $ and because $ X $ is poisson then $ Var(X) = E[X] $. but I'm not sure if $ Y $ is also a poisson variable?
Any suggestions?
probability poisson-distribution
probability poisson-distribution
edited Dec 9 '18 at 16:59
Nate Eldredge
63.6k682171
asked Dec 9 '18 at 16:37
bm1125bm1125
64916
edited Dec 9 '18 at 16:59
Nate Eldredge
63.6k682171
asked Dec 9 '18 at 16:37
bm1125bm1125
64916
edited Dec 9 '18 at 16:59
Nate Eldredge
63.6k682171
edited Dec 9 '18 at 16:59
Nate Eldredge
63.6k682171
edited Dec 9 '18 at 16:59
Nate Eldredge
63.6k682171
63.6k682171
asked Dec 9 '18 at 16:37
bm1125bm1125
64916
asked Dec 9 '18 at 16:37
bm1125bm1125
64916
asked Dec 9 '18 at 16:37
bm1125bm1125
64916
64916
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$Y$ cannot be Poisson, because for example, $$Pr[Y = 3] = Pr[2X = 3] = Pr[X = 3/2] = 0.$$
If we define $Y$ as in the body of the post, i.e. $$Y = begin{cases} 2X, & X le 3 \ X, & X ge 4, end{cases}$$ then we note that
$$begin{align*}operatorname{E}[Y^2]
&= sum_{x=0}^3 (2x)^2 e^{-lambda} frac{lambda^x}{x!} + sum_{x=4}^infty x^2 e^{-lambda} frac{lambda^x}{x!} \
&= 4 sum_{x=0}^3 x^2 e^{-lambda} frac{lambda^x}{x!} + sum_{x=4}^infty x^2 e^{-lambda} frac{lambda^x}{x!} \
&= 3 sum_{x=0}^3 x^2 e^{-lambda} frac{lambda^x}{x!} + left(sum_{x=0}^3 x^2 e^{-lambda} frac{lambda^x}{x!} + sum_{x=4}^infty x^2 e^{-lambda} frac{lambda^x}{x!} right) \
&= 3 sum_{x=0}^3 x^2 e^{-lambda} frac{lambda^x}{x!} + sum_{x=0}^infty x^2 e^{-lambda} frac{lambda^x}{x!} \
&= e^{-lambda} (3lambda + 6 lambda^2 + tfrac{9}{2} lambda^3) + operatorname{E}[X^2] \
&= 66e^{-2} + operatorname{E}[X]^2 + operatorname{Var}[X] \
&= 66e^{-2} + lambda^2 + lambda \
&= 6 + 66e^{-2}.
end{align*}$$
answered Dec 9 '18 at 16:54
heropupheropup
63.8k762102
$endgroup$
$begingroup$
Could you explain to me why $ (2x)^2 $ becomes $ 3x^2 $ ?
$endgroup$
– bm1125
Dec 9 '18 at 17:02
$begingroup$
@bm1125 I have expanded the details of that calculation in my edit, but simply put, $4 = 3 + 1.$
$endgroup$
– heropup
Dec 9 '18 at 17:07
add a comment |
$begingroup$
begin{align}
E[Y^2]
&= sum_{x=0}^3 (2x)^2 Pr(X=x)+ sum_{x=4}^infty x^2 Pr(X=x)\
&=sum_{x=0}^3 4x^2 Pr(X=x)+ sum_{x=4}^infty x^2 Pr(X=x)\
&=sum_{x=0}^3 3x^2 Pr(X=x)+ sum_{x=0}^infty x^2 Pr(X=x)\
&=3sum_{x=1}^3 x^2 Pr(X=x)+ E[X^2]\
&= 3left( Pr(X=1) + 4Pr(X=2)+9Pr(X=3) right) + E[X^2]
end{align}
answered Dec 9 '18 at 16:46
Siong Thye GohSiong Thye Goh
101k1466118
$endgroup$
$begingroup$
Thanks for your answer! could you explain the second equality? I think you also miss the fact that $ P{X=6} not = P{Y=6} $ as $ P{Y=6} = P{X=3} + P{X=6} $ I believe.
$endgroup$
– bm1125
Dec 9 '18 at 16:55
1
$begingroup$
All of the summations should be with respect to $x$.
$endgroup$
– heropup
Dec 9 '18 at 17:00
$begingroup$
oops, i made lots of typos, thanks.
$endgroup$
– Siong Thye Goh
Dec 9 '18 at 17:00
$begingroup$
I use the law of the unconscious statistician $E[g(X)]=sum_x g(x)P(X=x)$. For the line from $4x^$ to $3x^2$, we pass a copy of $x^2$ to the second term, hence there is change of the index.
$endgroup$
– Siong Thye Goh
Dec 9 '18 at 17:02
add a comment |
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2 Answers
2
active
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votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$Y$ cannot be Poisson, because for example, $$Pr[Y = 3] = Pr[2X = 3] = Pr[X = 3/2] = 0.$$
If we define $Y$ as in the body of the post, i.e. $$Y = begin{cases} 2X, & X le 3 \ X, & X ge 4, end{cases}$$ then we note that
$$begin{align*}operatorname{E}[Y^2]
&= sum_{x=0}^3 (2x)^2 e^{-lambda} frac{lambda^x}{x!} + sum_{x=4}^infty x^2 e^{-lambda} frac{lambda^x}{x!} \
&= 4 sum_{x=0}^3 x^2 e^{-lambda} frac{lambda^x}{x!} + sum_{x=4}^infty x^2 e^{-lambda} frac{lambda^x}{x!} \
&= 3 sum_{x=0}^3 x^2 e^{-lambda} frac{lambda^x}{x!} + left(sum_{x=0}^3 x^2 e^{-lambda} frac{lambda^x}{x!} + sum_{x=4}^infty x^2 e^{-lambda} frac{lambda^x}{x!} right) \
&= 3 sum_{x=0}^3 x^2 e^{-lambda} frac{lambda^x}{x!} + sum_{x=0}^infty x^2 e^{-lambda} frac{lambda^x}{x!} \
&= e^{-lambda} (3lambda + 6 lambda^2 + tfrac{9}{2} lambda^3) + operatorname{E}[X^2] \
&= 66e^{-2} + operatorname{E}[X]^2 + operatorname{Var}[X] \
&= 66e^{-2} + lambda^2 + lambda \
&= 6 + 66e^{-2}.
end{align*}$$
answered Dec 9 '18 at 16:54
heropupheropup
63.8k762102
$endgroup$
$begingroup$
Could you explain to me why $ (2x)^2 $ becomes $ 3x^2 $ ?
$endgroup$
– bm1125
Dec 9 '18 at 17:02
$begingroup$
@bm1125 I have expanded the details of that calculation in my edit, but simply put, $4 = 3 + 1.$
$endgroup$
– heropup
Dec 9 '18 at 17:07
add a comment |
$begingroup$
$Y$ cannot be Poisson, because for example, $$Pr[Y = 3] = Pr[2X = 3] = Pr[X = 3/2] = 0.$$
If we define $Y$ as in the body of the post, i.e. $$Y = begin{cases} 2X, & X le 3 \ X, & X ge 4, end{cases}$$ then we note that
$$begin{align*}operatorname{E}[Y^2]
&= sum_{x=0}^3 (2x)^2 e^{-lambda} frac{lambda^x}{x!} + sum_{x=4}^infty x^2 e^{-lambda} frac{lambda^x}{x!} \
&= 4 sum_{x=0}^3 x^2 e^{-lambda} frac{lambda^x}{x!} + sum_{x=4}^infty x^2 e^{-lambda} frac{lambda^x}{x!} \
&= 3 sum_{x=0}^3 x^2 e^{-lambda} frac{lambda^x}{x!} + left(sum_{x=0}^3 x^2 e^{-lambda} frac{lambda^x}{x!} + sum_{x=4}^infty x^2 e^{-lambda} frac{lambda^x}{x!} right) \
&= 3 sum_{x=0}^3 x^2 e^{-lambda} frac{lambda^x}{x!} + sum_{x=0}^infty x^2 e^{-lambda} frac{lambda^x}{x!} \
&= e^{-lambda} (3lambda + 6 lambda^2 + tfrac{9}{2} lambda^3) + operatorname{E}[X^2] \
&= 66e^{-2} + operatorname{E}[X]^2 + operatorname{Var}[X] \
&= 66e^{-2} + lambda^2 + lambda \
&= 6 + 66e^{-2}.
end{align*}$$
answered Dec 9 '18 at 16:54
heropupheropup
63.8k762102
$endgroup$
$begingroup$
Could you explain to me why $ (2x)^2 $ becomes $ 3x^2 $ ?
$endgroup$
– bm1125
Dec 9 '18 at 17:02
$begingroup$
@bm1125 I have expanded the details of that calculation in my edit, but simply put, $4 = 3 + 1.$
$endgroup$
– heropup
Dec 9 '18 at 17:07
add a comment |
$begingroup$
$Y$ cannot be Poisson, because for example, $$Pr[Y = 3] = Pr[2X = 3] = Pr[X = 3/2] = 0.$$
If we define $Y$ as in the body of the post, i.e. $$Y = begin{cases} 2X, & X le 3 \ X, & X ge 4, end{cases}$$ then we note that
$$begin{align*}operatorname{E}[Y^2]
&= sum_{x=0}^3 (2x)^2 e^{-lambda} frac{lambda^x}{x!} + sum_{x=4}^infty x^2 e^{-lambda} frac{lambda^x}{x!} \
&= 4 sum_{x=0}^3 x^2 e^{-lambda} frac{lambda^x}{x!} + sum_{x=4}^infty x^2 e^{-lambda} frac{lambda^x}{x!} \
&= 3 sum_{x=0}^3 x^2 e^{-lambda} frac{lambda^x}{x!} + left(sum_{x=0}^3 x^2 e^{-lambda} frac{lambda^x}{x!} + sum_{x=4}^infty x^2 e^{-lambda} frac{lambda^x}{x!} right) \
&= 3 sum_{x=0}^3 x^2 e^{-lambda} frac{lambda^x}{x!} + sum_{x=0}^infty x^2 e^{-lambda} frac{lambda^x}{x!} \
&= e^{-lambda} (3lambda + 6 lambda^2 + tfrac{9}{2} lambda^3) + operatorname{E}[X^2] \
&= 66e^{-2} + operatorname{E}[X]^2 + operatorname{Var}[X] \
&= 66e^{-2} + lambda^2 + lambda \
&= 6 + 66e^{-2}.
end{align*}$$
answered Dec 9 '18 at 16:54
heropupheropup
63.8k762102
$endgroup$
$Y$ cannot be Poisson, because for example, $$Pr[Y = 3] = Pr[2X = 3] = Pr[X = 3/2] = 0.$$
If we define $Y$ as in the body of the post, i.e. $$Y = begin{cases} 2X, & X le 3 \ X, & X ge 4, end{cases}$$ then we note that
$$begin{align*}operatorname{E}[Y^2]
&= sum_{x=0}^3 (2x)^2 e^{-lambda} frac{lambda^x}{x!} + sum_{x=4}^infty x^2 e^{-lambda} frac{lambda^x}{x!} \
&= 4 sum_{x=0}^3 x^2 e^{-lambda} frac{lambda^x}{x!} + sum_{x=4}^infty x^2 e^{-lambda} frac{lambda^x}{x!} \
&= 3 sum_{x=0}^3 x^2 e^{-lambda} frac{lambda^x}{x!} + left(sum_{x=0}^3 x^2 e^{-lambda} frac{lambda^x}{x!} + sum_{x=4}^infty x^2 e^{-lambda} frac{lambda^x}{x!} right) \
&= 3 sum_{x=0}^3 x^2 e^{-lambda} frac{lambda^x}{x!} + sum_{x=0}^infty x^2 e^{-lambda} frac{lambda^x}{x!} \
&= e^{-lambda} (3lambda + 6 lambda^2 + tfrac{9}{2} lambda^3) + operatorname{E}[X^2] \
&= 66e^{-2} + operatorname{E}[X]^2 + operatorname{Var}[X] \
&= 66e^{-2} + lambda^2 + lambda \
&= 6 + 66e^{-2}.
end{align*}$$
answered Dec 9 '18 at 16:54
heropupheropup
63.8k762102
edited Dec 9 '18 at 17:06
answered Dec 9 '18 at 16:54
heropupheropup
63.8k762102
answered Dec 9 '18 at 16:54
heropupheropup
63.8k762102
answered Dec 9 '18 at 16:54
heropupheropup
63.8k762102
63.8k762102
$begingroup$
Could you explain to me why $ (2x)^2 $ becomes $ 3x^2 $ ?
$endgroup$
– bm1125
Dec 9 '18 at 17:02
$begingroup$
@bm1125 I have expanded the details of that calculation in my edit, but simply put, $4 = 3 + 1.$
$endgroup$
– heropup
Dec 9 '18 at 17:07
add a comment |
$begingroup$
Could you explain to me why $ (2x)^2 $ becomes $ 3x^2 $ ?
$endgroup$
– bm1125
Dec 9 '18 at 17:02
$begingroup$
@bm1125 I have expanded the details of that calculation in my edit, but simply put, $4 = 3 + 1.$
$endgroup$
– heropup
Dec 9 '18 at 17:07
$begingroup$
Could you explain to me why $ (2x)^2 $ becomes $ 3x^2 $ ?
$endgroup$
– bm1125
Dec 9 '18 at 17:02
$begingroup$
Could you explain to me why $ (2x)^2 $ becomes $ 3x^2 $ ?
$endgroup$
– bm1125
Dec 9 '18 at 17:02
$begingroup$
@bm1125 I have expanded the details of that calculation in my edit, but simply put, $4 = 3 + 1.$
$endgroup$
– heropup
Dec 9 '18 at 17:07
$begingroup$
@bm1125 I have expanded the details of that calculation in my edit, but simply put, $4 = 3 + 1.$
$endgroup$
– heropup
Dec 9 '18 at 17:07
add a comment |
$begingroup$
begin{align}
E[Y^2]
&= sum_{x=0}^3 (2x)^2 Pr(X=x)+ sum_{x=4}^infty x^2 Pr(X=x)\
&=sum_{x=0}^3 4x^2 Pr(X=x)+ sum_{x=4}^infty x^2 Pr(X=x)\
&=sum_{x=0}^3 3x^2 Pr(X=x)+ sum_{x=0}^infty x^2 Pr(X=x)\
&=3sum_{x=1}^3 x^2 Pr(X=x)+ E[X^2]\
&= 3left( Pr(X=1) + 4Pr(X=2)+9Pr(X=3) right) + E[X^2]
end{align}
answered Dec 9 '18 at 16:46
Siong Thye GohSiong Thye Goh
101k1466118
$endgroup$
$begingroup$
Thanks for your answer! could you explain the second equality? I think you also miss the fact that $ P{X=6} not = P{Y=6} $ as $ P{Y=6} = P{X=3} + P{X=6} $ I believe.
$endgroup$
– bm1125
Dec 9 '18 at 16:55
1
$begingroup$
All of the summations should be with respect to $x$.
$endgroup$
– heropup
Dec 9 '18 at 17:00
$begingroup$
oops, i made lots of typos, thanks.
$endgroup$
– Siong Thye Goh
Dec 9 '18 at 17:00
$begingroup$
I use the law of the unconscious statistician $E[g(X)]=sum_x g(x)P(X=x)$. For the line from $4x^$ to $3x^2$, we pass a copy of $x^2$ to the second term, hence there is change of the index.
$endgroup$
– Siong Thye Goh
Dec 9 '18 at 17:02
add a comment |
$begingroup$
begin{align}
E[Y^2]
&= sum_{x=0}^3 (2x)^2 Pr(X=x)+ sum_{x=4}^infty x^2 Pr(X=x)\
&=sum_{x=0}^3 4x^2 Pr(X=x)+ sum_{x=4}^infty x^2 Pr(X=x)\
&=sum_{x=0}^3 3x^2 Pr(X=x)+ sum_{x=0}^infty x^2 Pr(X=x)\
&=3sum_{x=1}^3 x^2 Pr(X=x)+ E[X^2]\
&= 3left( Pr(X=1) + 4Pr(X=2)+9Pr(X=3) right) + E[X^2]
end{align}
answered Dec 9 '18 at 16:46
Siong Thye GohSiong Thye Goh
101k1466118
$endgroup$
$begingroup$
Thanks for your answer! could you explain the second equality? I think you also miss the fact that $ P{X=6} not = P{Y=6} $ as $ P{Y=6} = P{X=3} + P{X=6} $ I believe.
$endgroup$
– bm1125
Dec 9 '18 at 16:55
1
$begingroup$
All of the summations should be with respect to $x$.
$endgroup$
– heropup
Dec 9 '18 at 17:00
$begingroup$
oops, i made lots of typos, thanks.
$endgroup$
– Siong Thye Goh
Dec 9 '18 at 17:00
$begingroup$
I use the law of the unconscious statistician $E[g(X)]=sum_x g(x)P(X=x)$. For the line from $4x^$ to $3x^2$, we pass a copy of $x^2$ to the second term, hence there is change of the index.
$endgroup$
– Siong Thye Goh
Dec 9 '18 at 17:02
add a comment |
$begingroup$
begin{align}
E[Y^2]
&= sum_{x=0}^3 (2x)^2 Pr(X=x)+ sum_{x=4}^infty x^2 Pr(X=x)\
&=sum_{x=0}^3 4x^2 Pr(X=x)+ sum_{x=4}^infty x^2 Pr(X=x)\
&=sum_{x=0}^3 3x^2 Pr(X=x)+ sum_{x=0}^infty x^2 Pr(X=x)\
&=3sum_{x=1}^3 x^2 Pr(X=x)+ E[X^2]\
&= 3left( Pr(X=1) + 4Pr(X=2)+9Pr(X=3) right) + E[X^2]
end{align}
answered Dec 9 '18 at 16:46
Siong Thye GohSiong Thye Goh
101k1466118
$endgroup$
begin{align}
E[Y^2]
&= sum_{x=0}^3 (2x)^2 Pr(X=x)+ sum_{x=4}^infty x^2 Pr(X=x)\
&=sum_{x=0}^3 4x^2 Pr(X=x)+ sum_{x=4}^infty x^2 Pr(X=x)\
&=sum_{x=0}^3 3x^2 Pr(X=x)+ sum_{x=0}^infty x^2 Pr(X=x)\
&=3sum_{x=1}^3 x^2 Pr(X=x)+ E[X^2]\
&= 3left( Pr(X=1) + 4Pr(X=2)+9Pr(X=3) right) + E[X^2]
end{align}
answered Dec 9 '18 at 16:46
Siong Thye GohSiong Thye Goh
101k1466118
edited Dec 9 '18 at 17:00
answered Dec 9 '18 at 16:46
Siong Thye GohSiong Thye Goh
101k1466118
answered Dec 9 '18 at 16:46
Siong Thye GohSiong Thye Goh
101k1466118
answered Dec 9 '18 at 16:46
Siong Thye GohSiong Thye Goh
101k1466118
101k1466118
$begingroup$
Thanks for your answer! could you explain the second equality? I think you also miss the fact that $ P{X=6} not = P{Y=6} $ as $ P{Y=6} = P{X=3} + P{X=6} $ I believe.
$endgroup$
– bm1125
Dec 9 '18 at 16:55
1
$begingroup$
All of the summations should be with respect to $x$.
$endgroup$
– heropup
Dec 9 '18 at 17:00
$begingroup$
oops, i made lots of typos, thanks.
$endgroup$
– Siong Thye Goh
Dec 9 '18 at 17:00
$begingroup$
I use the law of the unconscious statistician $E[g(X)]=sum_x g(x)P(X=x)$. For the line from $4x^$ to $3x^2$, we pass a copy of $x^2$ to the second term, hence there is change of the index.
$endgroup$
– Siong Thye Goh
Dec 9 '18 at 17:02
add a comment |
$begingroup$
Thanks for your answer! could you explain the second equality? I think you also miss the fact that $ P{X=6} not = P{Y=6} $ as $ P{Y=6} = P{X=3} + P{X=6} $ I believe.
$endgroup$
– bm1125
Dec 9 '18 at 16:55
1
$begingroup$
All of the summations should be with respect to $x$.
$endgroup$
– heropup
Dec 9 '18 at 17:00
$begingroup$
oops, i made lots of typos, thanks.
$endgroup$
– Siong Thye Goh
Dec 9 '18 at 17:00
$begingroup$
I use the law of the unconscious statistician $E[g(X)]=sum_x g(x)P(X=x)$. For the line from $4x^$ to $3x^2$, we pass a copy of $x^2$ to the second term, hence there is change of the index.
$endgroup$
– Siong Thye Goh
Dec 9 '18 at 17:02
$begingroup$
Thanks for your answer! could you explain the second equality? I think you also miss the fact that $ P{X=6} not = P{Y=6} $ as $ P{Y=6} = P{X=3} + P{X=6} $ I believe.
$endgroup$
– bm1125
Dec 9 '18 at 16:55
$begingroup$
Thanks for your answer! could you explain the second equality? I think you also miss the fact that $ P{X=6} not = P{Y=6} $ as $ P{Y=6} = P{X=3} + P{X=6} $ I believe.
$endgroup$
– bm1125
Dec 9 '18 at 16:55
1
1
$begingroup$
All of the summations should be with respect to $x$.
$endgroup$
– heropup
Dec 9 '18 at 17:00
$begingroup$
All of the summations should be with respect to $x$.
$endgroup$
– heropup
Dec 9 '18 at 17:00
$begingroup$
oops, i made lots of typos, thanks.
$endgroup$
– Siong Thye Goh
Dec 9 '18 at 17:00
$begingroup$
oops, i made lots of typos, thanks.
$endgroup$
– Siong Thye Goh
Dec 9 '18 at 17:00
$begingroup$
I use the law of the unconscious statistician $E[g(X)]=sum_x g(x)P(X=x)$. For the line from $4x^$ to $3x^2$, we pass a copy of $x^2$ to the second term, hence there is change of the index.
$endgroup$
– Siong Thye Goh
Dec 9 '18 at 17:02
$begingroup$
I use the law of the unconscious statistician $E[g(X)]=sum_x g(x)P(X=x)$. For the line from $4x^$ to $3x^2$, we pass a copy of $x^2$ to the second term, hence there is change of the index.
$endgroup$
– Siong Thye Goh
Dec 9 '18 at 17:02
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