Find the distribution of the random variable $Y=\sum _{j=0}^{d\:}\:X^2_j$ if $ X^2_1,\:X^2_2,\:.....,\:X^2_d $ are all independent and distributed with a chi-squared distribution with $ v=1 $ degree of freedom, Say, $ X^2_j\:$~$\:X^2_1,\:j=1,.....,d $

How do I solve this question ?


I guess you mean that $\mathrm{X}_j^2\sim \chi_1^2$.

The $\chi_n^2$ distribution is defined as:


where $X_i\sim N(0,1)$. So then $\chi_1^2\sim X^2$ where $X\sim N(0,1)$.

You are asked to find the distribution of $Y=\sum_{j=1}^dX_i^2$ which follows the same distribution as

$Y=\sum_{j=1}^dX_i^2$ where $X_i\sim N(0,1)$, for all $j=1,...,d$.

Applying the definition above, you get that because of the construction of the $\chi_n^2$:

$Y\sim \chi_d^2$

that is, a chi-squared distribuion with d degrees of freedom.

  • $\begingroup$ yes sorry that's what I meant, I didn't get what you mean starting from the 5th line,, can you please explain the steps you did to get the answer $\endgroup$ – ASD123 Oct 26 '15 at 23:49
  • $\begingroup$ @ASD123 Okay. You need to sum d random variables, each one following a chi-squared distribution with one degree of freedom. But a chi squared distribution with one degree of freedom is just the square of a random variable X following a N(0,1) distribution (you can check that pluggin n=1 in the formula on the second line). Then, the sum of d chi-squared distributed r-v's is equal to the sum of d X^2 random variables following a N(0,1) distribution, which by the formula on the third line, is equal to a chi-squared distribution with d degrees of freedom. $\endgroup$ – user281593 Oct 27 '15 at 0:09

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