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Given $X_1 \sim \mathcal{N}(\mu_1,\sigma_1^2)$ and $X_2 \sim \mathcal{N}(\mu_2,\sigma_2^2)$, with $X_1$ independent of $X_2$, as well as $Y = (X_1 + X_2)^2$, what is $\Pr(Y>\mathrm{E}[Y])$?

Misguided attempt: $$ \begin{align} &Y \sim \left(\mathcal{N}(\mu=1,\sigma^2=5^2)\right)^2\\ &\Pr(Y>\mathrm{E}[Y]) = \int_{\mathrm{E}[Y]}^\infty f(y) dy\\ &\Pr(Y>\mathrm{E}[Y]) = \int_{\mathrm{E}[Y]}^\infty\left( \frac{1}{\sigma \sqrt{2\pi}}\exp\left(\frac{-(x-\mu)^2}{2\sigma^2}\right)\right)^2dx\\ &\Pr(Y>\mathrm{E}[Y]) = \frac{1}{2\sigma^2\pi}\int_{\mathrm{E}[Y]}^\infty\exp\left(\frac{-(x-\mu)^2}{\sigma^2}\right)dx\\ &\Pr(Y>\mathrm{E}[Y]) = \text{... don't know...} \end{align} $$ Insight into where the thinking is misguided, as well as what the right approach is, would be appreciated.

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We know that if $X_1$ and $X_2$ are independent normal random variables, then their sum $$X_1 + X_2 \sim \operatorname{Normal}(\mu_1 + \mu_2, \sigma_1^2 + \sigma_2^2).$$ Therefore, $$\mu_Y = \operatorname{E}[Y] = \operatorname{E}[(X_1 + X_2)^2] = \operatorname{Var}[X_1 + X_2] + \operatorname{E}[X_1+X_2]^2 = \sigma_1^2 + \sigma_2^2 + (\mu_1 + \mu_2)^2.$$ Note that this expectation is strictly positive. Now consider the probability $$\Pr[Y > \mu_Y] = \Pr[X_1 + X_2 > \sqrt{\mu_Y}] + \Pr[X_1 + X_2 < -\sqrt{\mu_Y}],$$ which is a direct consequence of applying the inverse transformation (take the square root of both sides, recalling that $\mu_Y > 0$). But we already established that $X_1 + X_2$ is normally distributed, so we simply standardize and write $$\Pr[X_1 + X_2 > \sqrt{\mu_Y}] = \Pr\left[\frac{X_1 + X_2 - (\mu_1 + \mu_2)}{\sqrt{\sigma_1^2 + \sigma_2^2}} > \frac{\sqrt{\mu_Y} - (\mu_1 + \mu_2)}{\sqrt{\sigma_1^2 + \sigma_2^2}}\right],$$ and the LHS of the inequality is a standard normal variable $Z$. A similar approach applies to the second term above. Thus $$\Pr[Y > \mu_Y] = 1 - \Phi\left(\frac{\sqrt{\mu_Y} - (\mu_1 + \mu_2)}{\sqrt{\sigma_1^2 + \sigma_2^2}}\right) + \Phi\left(\frac{-\sqrt{\mu_Y} - (\mu_1 + \mu_2)}{\sqrt{\sigma_1^2 + \sigma_2^2}}\right),$$ where $\Phi$ is the cumulative distribution of the standard normal distribution, and $\mu_Y$ is defined as above.

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  • $\begingroup$ Excellent, thank you @heropup. Any insight into why the above attempt was misguided? $\endgroup$
    – Arthur D.
    Feb 5, 2016 at 13:44
  • $\begingroup$ @ArthurD. If $Y = g(X)$ for some function $g$ and some random variable $X$, then that does not imply $f_Y(y) = g(f_X(y))$. $\endgroup$
    – heropup
    Feb 5, 2016 at 16:18

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