The kurtosis of standard normal pdf

[pasivisku]

Hi, I'm tying to solve the fourth moment of pdf of standard normal pdf. meanwhile, i have to break this one:
how to get the result of
integral of x exp (x) dx.

that would mean a great difference to me. thank you.

 

[pasivisku]

sorry to bother but problem solved...
I'll come to y'all next time in future
bye.

 

[tacman]

Hello,

The kurtosis of a standard normal probability density function (pdf) is equal to 3. This is because the standard normal distribution is a symmetric bell-shaped curve with a mean of 0 and a standard deviation of 1, which results in a kurtosis of 3.

To solve for the fourth moment of the pdf of a standard normal distribution, you can use the formula for kurtosis which is:

Kurtosis = (1/σ^4) * ∫(x-μ)^4 * f(x) dx

where σ is the standard deviation and μ is the mean of the distribution.

Since the standard normal distribution has a mean of 0 and a standard deviation of 1, the formula simplifies to:

Kurtosis = ∫x^4 * f(x) dx

To solve for the integral of x * exp(x) dx, you can use integration by parts. Let u = x and dv = exp(x) dx. Then, du = dx and v = exp(x). Using the formula for integration by parts, we get:

∫x * exp(x) dx = x * exp(x) - ∫exp(x) dx

= x * exp(x) - exp(x) + C

= (x - 1) * exp(x) + C

Therefore, the result of the integral is (x - 1) * exp(x) + C.

I hope this helps. Let me know if you have any further questions. Thank you.