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The question is not well defined. The best way to "do something" is to "guess" the or a related question, and answer this one. (The original question must have been in the same circle of ideas, and should have been touched with similar vehicles.)

Restatement:

Let us fix an integer $K>1$. We consider

• $K$ random variables $Z_1,\dots, Z_K$ which follow the standard normal distribution $N(0,1^2)$,

• and $K$ random variables $V_1,\dots, V_K$ which follow the uniform distribution on the intervals $(1,2),\dots,(1,K+1)$ - respectively.

The family of all these variables should be an independent family of random variables defined on the same probability space. Let $\mathbb{E}$ be the expectation, the mean on this space. We build $X(K)=|Z_1V_1+\dots+Z_KV_K|$ and its expectation $f(K)= \mathbb{E} X(K)=\mathbb{E}\Big[\ |Z_1V_1+\dots+Z_KV_K|\ \Big]$ as a function of $K$. The exercise asks for

• heuristical arguments, that may lead to an asymptotic $F(K)=O(K^?)$ in big-O-notation, and

• a computer simulation that supports the heuristic.

This was the complicated part of the answer. From this point things go straightforward: The random variable under the modulus has mean zero since $ \mathbb{E} [Z_jV_j] = \mathbb{E} [Z_j] \mathbb{E} [V_j] = 0\cdot \mathbb{E} [V_j]=0$, and terms have variance Var$[Z_jV_j] = \mathbb{E} [(Z_jV_j)^2] -\mathbb{E} [Z_jV_j]^2 =\mathbb{E} [Z_j^2] \mathbb{E} [V_j^2] -\mathbb{E} [Z_j]\mathbb{E} [V_j]^2 = 1\cdot \mathbb{E} [V_j^2]-0 =\frac 1{j}\int_1^{j+1}x^2\; dv=\frac 1{3j}((j+1)^3-1^3)$ and so on. (We used independence. Further using the independence, the variance of the sum is the sum of the variances and we compute $\sum_{1\le j\le K}\frac 1{3j}((j+1)^3-1^3)$:

sage: var( 'j,K' );
sage: latex( sum( 1/3/j * ( (j+1)^3-1^3 ), j, 1, K ).factor() )
\frac{1}{9} \, {\left(K^{2} + 6 \, K + 14\right)} K

Then we expect: $\displaystyle \frac{Z_1V_1+\dots+Z_KV_K}{\displaystyle\sqrt{\frac{1}{9} {\left(K^{2} + 6 \, K + 14\right)} K}} \sim N(0,1^2)$ . (This is the optimistic law of large numbers, used outside mathematics when we do not have time to check the details.) For a big $K$ we can optimistically and statistically approximate the RHS with a normally distributed $Y\in N(0,1^2)$. Then $\mathbb{E}|Y|$ is twice the integral on $[0,\infty)$ from $\frac1 {\sqrt{2\pi}}y\exp(-y^2/2)$. Putting all together we get: $\displaystyle f(K)\sim \frac 2{3\sqrt {2\pi}} \sqrt{K\left(K^{2} + 6 K + 14\right)}$ .

That's the maths.

Now we simulate and we ask also for the values respecting the guessed asymptotic:

• The simulateion:

  sage: %cpaste Pasting code; enter '--' alone on the line to stop or use Ctrl-D. : :for pow in [ 2,3,4,5 ]: : K = 10 ** pow : SAMPLES = [] # and we append : for experiment in [ 1..99 ]: : SAMPLES . append( abs( sum( [ gauss(0,1) * uniform( 1,k+2 ) for k in range(K) ] ) ) ) : print "%s -> %s" % ( K, mean( SAMPLES ) ) :-- 100 -> 294.711735785 1000 -> 8714.8222098 10000 -> 249403.620665 100000 -> 8734793.09067

Next time we will see other numbers above.

• And the asymptotic: sage: for pow in [ 2,3,4,5 ]: K = 10 * pow print "%s -> %f" % ( K, 2/3/sqrt(2pi) * sqrt( K * ( K^2 + 6*K + 14 ) ) ) ....:
  100 -> 274.004912 1000 -> 8435.694028 10000 -> 266041.315371 100000 -> 8410694.055422

I did not check the details, but we strongly encourage $f(K)\in O(K^{3/2})$, even more, we have $f(K)\sim\frac 1{\sqrt{2\pi}}\cdot\frac 23\cdot K^{3/2}$.

The question is not well defined. The best way to "do something" is to "guess" the or a related question, and answer this one. (The original question must have been in the same circle of ideas, and should have been touched with similar vehicles.)

Restatement:

Let us fix an integer $K>1$. We consider

• $K$ random variables $Z_1,\dots, Z_K$ which follow the standard normal distribution $N(0,1^2)$,

• and $K$ random variables $V_1,\dots, V_K$ which follow the uniform distribution on the intervals $(1,2),\dots,(1,K+1)$ - respectively.

The family of all these variables should be an independent family of random variables defined on the same probability space. Let $\mathbb{E}$ be the expectation, the mean on this space. We build $X(K)=|Z_1V_1+\dots+Z_KV_K|$ and its expectation $f(K)= \mathbb{E} X(K)=\mathbb{E}\Big[\ |Z_1V_1+\dots+Z_KV_K|\ \Big]$ as a function of $K$. The exercise asks for

• heuristical arguments, that may lead to an asymptotic $F(K)=O(K^?)$ in big-O-notation, and

• a computer simulation that supports the heuristic.

This was the complicated part of the answer. From this point things go straightforward: The random variable under the modulus has mean zero since $ \mathbb{E} [Z_jV_j] = \mathbb{E} [Z_j] \mathbb{E} [V_j] = 0\cdot \mathbb{E} [V_j]=0$, and terms have variance Var$[Z_jV_j]

Var$\displaystyle[Z_jV_j] = \mathbb{E} [(Z_jV_j)^2] -\mathbb{E} [Z_jV_j]^2 =\mathbb{E} [Z_j^2] \mathbb{E} [V_j^2] -\mathbb{E} [Z_j]\mathbb{E} [V_j]^2 [V_j]^2$

$\qquad\displaystyle = 1\cdot \mathbb{E} [V_j^2]-0 =\frac 1{j}\int_1^{j+1}x^2\; dv=\frac 1{3j}((j+1)^3-1^3)$

and so on. (We

We used independence. Further using the independence, the variance of the sum is the sum of the variances and we compute $\sum_{1\le $\displaystyle\sum_{1\le j\le K}\frac 1{3j}((j+1)^3-1^3)$:

sage: var( 'j,K' );
sage: latex( sum( 1/3/j * ( (j+1)^3-1^3 ), j, 1, K ).factor() )
\frac{1}{9} \, {\left(K^{2} + 6 \, K + 14\right)} K

Then we expect: $\displaystyle \frac{Z_1V_1+\dots+Z_KV_K}{\displaystyle\sqrt{\frac{1}{9} {\left(K^{2} + 6 \, K + 14\right)} K}} \sim N(0,1^2)$ .

(This is the optimistic law of large numbers, used applied outside mathematics when we do not have time to check the details.)

For a big $K$ we can optimistically and statistically approximate the RHS with a normally distributed $Y\in N(0,1^2)$.

Then $\mathbb{E}|Y|$ is twice the integral on $[0,\infty)$ from $\frac1 {\sqrt{2\pi}}y\exp(-y^2/2)$. {\sqrt{2\pi}}y\exp(-y^2/2)$.

Putting all together we get: $\displaystyle f(K)\sim \frac 2{3\sqrt {2\pi}} \sqrt{K\left(K^{2} + 6 K + 14\right)}$ .

That's the maths.

Now we simulate and we ask also for the values respecting the guessed asymptotic:

• The simulateion:

  sage: %cpaste Pasting code; enter '--' alone on the line to stop or use Ctrl-D. : :for simulation:

  for pow in [ 2,3,4,5 ]: : K = 10 ** pow : SAMPLES = [] # and we append : for experiment in [ 1..99 ]: : SAMPLES . append( abs( sum( [ gauss(0,1) * uniform( 1,k+2 ) for k in range(K) ] ) ) ) : print "%s -> %s" % ( K, mean( SAMPLES ) ) :-- )

  100 -> 294.711735785 1000 -> 8714.8222098 10000 -> 249403.620665 100000 -> 8734793.09067

Next time we will see other numbers above.

• And the asymptotic: asymptotic:

  sage: for pow in [ 2,3,4,5 ]: K = 10 * pow print "%s -> %f" % ( K, 2/3/sqrt(2pi) * sqrt( K * ( K^2 + 6*K + 14 ) ) ) ....:
  )

  100 -> 274.004912 1000 -> 8435.694028 10000 -> 266041.315371 100000 -> 8410694.055422

I did not check the details, but we strongly encourage $f(K)\in O(K^{3/2})$, even more, we have $f(K)\sim\frac

$\displaystyle f(K)\sim\frac 1{\sqrt{2\pi}}\cdot\frac 23\cdot K^{3/2}$.

The question is not well defined. The best way to "do something" is to "guess" the or a related question, and answer this one. (The original question must have been in the same circle of ideas, and should have been touched with similar vehicles.)

Restatement:

Let us fix an integer $K>1$. We consider

• $K$ random variables $Z_1,\dots, Z_K$ which follow the standard normal distribution $N(0,1^2)$,

• and $K$ random variables $V_1,\dots, V_K$ which follow the uniform distribution on the intervals $(1,2),\dots,(1,K+1)$ - respectively.

The family of all these variables should be an independent family of random variables defined on the same probability space. Let $\mathbb{E}$ be the expectation, the mean on this space. We build $X(K)=|Z_1V_1+\dots+Z_KV_K|$ and its expectation $f(K)= \mathbb{E} X(K)=\mathbb{E}\Big[\ |Z_1V_1+\dots+Z_KV_K|\ \Big]$ as a function of $K$. The exercise asks for

• heuristical arguments, that may lead to an asymptotic $F(K)=O(K^?)$ in big-O-notation, and

• a computer simulation that supports the heuristic.

This was the complicated part of the answer. From this point things go straightforward: The random variable under the modulus has mean zero since $ \mathbb{E} [Z_jV_j] = \mathbb{E} [Z_j] \mathbb{E} [V_j] = 0\cdot \mathbb{E} [V_j]=0$, and terms have variance

Var$\displaystyle[Z_jV_j] = \mathbb{E} [(Z_jV_j)^2] -\mathbb{E} [Z_jV_j]^2 =\mathbb{E} [Z_j^2] \mathbb{E} [V_j^2] -\mathbb{E} [Z_j]\mathbb{E} [V_j]^2$

$\qquad\displaystyle = 1\cdot \mathbb{E} [V_j^2]-0 =\frac 1{j}\int_1^{j+1}x^2\; dv=\frac 1{3j}((j+1)^3-1^3)$

and so on.

We used independence. Further using the independence, the variance of the sum is the sum of the variances and we compute $\displaystyle\sum_{1\le j\le K}\frac 1{3j}((j+1)^3-1^3)$:

sage: var( 'j,K' );
sage: latex( sum( 1/3/j * ( (j+1)^3-1^3 ), j, 1, K ).factor() )
\frac{1}{9} \, {\left(K^{2} + 6 \, K + 14\right)} K

Then we expect: $\displaystyle \frac{Z_1V_1+\dots+Z_KV_K}{\displaystyle\sqrt{\frac{1}{9} {\left(K^{2} + 6 \, K + 14\right)} K}} \sim N(0,1^2)$ .

(This is the optimistic law of large numbers, applied outside mathematics when we do not have time to check the details.)

For a big $K$ we can optimistically and statistically approximate the RHS with a normally distributed $Y\in N(0,1^2)$.

Then $\mathbb{E}|Y|$ is twice the integral on $[0,\infty)$ from $\frac1 {\sqrt{2\pi}}y\exp(-y^2/2)$.

Putting all together we get: $\displaystyle f(K)\sim \frac 2{3\sqrt {2\pi}} \sqrt{K\left(K^{2} + 6 K + 14\right)}$ .

That's the maths.

Now we simulate and we ask also for the values respecting the guessed asymptotic:

Next time we will see other numbers above.

I did not check the details, but we strongly encourage $f(K)\in O(K^{3/2})$, even more, we have

$\displaystyle f(K)\sim\frac 1{\sqrt{2\pi}}\cdot\frac 23\cdot K^{3/2}$.