The Distribution of $T = (X_1 - \mu)^2 + (X_2 - \mu)^2 + \cdots + (X_{10} - \mu)^2$ I am having a little bit of trouble obtaining the distribution for $T = (X_1 - \mu)^2 + (X_2 - \mu)^2 + \cdots + (X_{10} - \mu)^2$, where $X_1,...,X_{10}$ are independent and identically distributed normal with mean $\mu$ and variance 1.
I know that the normal distribution with mean $\mu$ and variance 1 is defined by
\begin{equation*}
f(x_i) = \dfrac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}\left(x_i - \mu\right)^2}
\end{equation*}
So if $T = (X_1 - \mu)^2 + (X_2 - \mu)^2 + \cdots + (X_{10} - \mu)^2$, then would the distribution turn out to be
\begin{align*}
f(t) = \dfrac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}\left[\left(x_1 - \mu\right)^2 - \mu\right]^2} + \dfrac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}\left[\left(x_2 - \mu\right)^2 - \mu\right]^2} + \cdots + \dfrac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}\left[\left(x_{10} - \mu\right)^2 - \mu\right]^2} =\dfrac{1}{\sqrt{2\pi}}\sum_{i = 1}^{10} e^{-\frac{1}{2}\left[\left(x_i - \mu\right)^2 - \mu\right]^2}
\end{align*}
I am not sure if this is the correct answer to this. Would appreciate some assistance.
 A: So first, if $X$ is a normal random variable with mean $\mu$ and variance $1$ I will use the notation$ X \sim \mathcal{N}(\mu,1) $ lets figure out the distribution of $X-\mu$. The distribution of $X$ is:
$$
\mathbb{P}(X \leq x) = \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}(x-\mu)^2}
$$
But the distribution of $X -\mu$ :
$$
\mathbb{P}(X-\mu \leq x) = \mathbb{P}(X \leq x +\mu) =  \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}
$$
Which means that $X + \mu \sim \mathcal{N}(0,1)$. That is, if we substract from X the constant $\mu$ we obtain the a normally distributed random variable with mean $1$ and variance $0$.
Now we need to find the distribution of the square of a normally distributed random variable, I will write the result, but for a detailed part of this step you can read this detailed explanation, specifically I recommend the "CDF technique" :
$$
\mathbb{P}(X^2 \leq x) = \frac{1}{\sqrt{x}} \frac{1}{\sqrt{2\pi}}e^{-x/2} 
$$
Which is the chi-squared distribution. So far, we have found that $(X-\mu)^2$ has $\chi(1)$ distribution with one degree of freedom. Now we just need to find the distribution of the sum of $\chi(1)$ random variables. It turns out that the sum of $n$ distributed $\chi(1)$ random variables is a $\chi(n)$ random variable. This is, you need to sum their degrees of freedom. So in your case we obtain that $T \sim \chi(10) $
