1
$\begingroup$

Suppose, $X_1,X_2,X_3,X_4$ are i.i.d. random variables with values 1 and -1 with prob 0.5 each. Then find the value of $E(X_1+X_2+X_3+X_4)^4$.

Ans:

Let, $Y=\sum_{1}^{4}X_{i}$ Now, Y takes values 4,-4,2,-2,0 with the following probabilities:

$P(Y=0)=\frac{6}{2^4}$

$P(Y=2)=\frac{4}{2^4}=P(Y=-2)$

$P(Y=4)=\frac{1}{2^4}=P(Y=-4)$

$E(Y^4)=2^4 \frac{4}{2^4}+(-2)^4 \frac{4}{2^4}+4^4 \frac{1}{2^4}+(-4)^4 \frac{1}{2^4}=4+4+16+16=40$

Is it correct?

$\endgroup$
  • 1
    $\begingroup$ I presume the power is applied before the expectation? $\endgroup$ – Denziloe Aug 25 '18 at 16:30
  • 2
    $\begingroup$ What specific part of the solution do you want help with? Please note that we don't field yes/no questions here: we're looking for more substance than that. $\endgroup$ – whuber Aug 25 '18 at 17:53
0
$\begingroup$

When in doubt I like to double check my work by running a simulation with Python. I sampled uniformly from the two possible values of -1 and 1 and created an array with $10^7$ rows and 4 columns.

array([[-1, -1,  1,  1],
       [ 1, -1, -1,  1],
       [-1, -1,  1, -1],
       ...,
       [ 1,  1,  1, -1],
       [ 1,  1,  1, -1],
       [ 1,  1,  1,  1]])

The columns represent the $X_i$ values for a random trial and the sum of each row represents the $Y$ variables. Raising the sum of each row to the fourth power and then taking the mean yields a value that is approximately 40.

import numpy as np

possible_values = (-1, 1)
n_trials = 10**7
Xi_values = np.random.choice(possible_values, size=n_trials*4).reshape(n_trials, 4)
Y_values = np.sum(Xi_values, axis=1)
np.mean(Y_values**4)
40.0041056
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.