suppose $X$ and $Y$ are independent random variables with standard normal distributions. The probability of $X<-1$ is some p which lies in the open interval (0,1). what is the probability of the event: $X^2 > 1$ and $Y^3 > 1$.

I have started with $P(X<-1)= P(X>1) = p$ (i think its right) But now stuck at how to proceed from here... I'm thinking of using cdf but not able to fit it in?


$$P((X^2 > 1) \cap Y^3 > 1)$$ $$=P(X^2 > 1)\times P(Y > 1)$$ $$=P(X > 1)\times P(Y>1) + P(X < -1)\times P(Y>1)$$

$$=(1-\Phi(1))(1-\Phi(1)) + \Phi(-1)\times (1-\Phi(1))$$

where $\Phi(.)$ denotes the CDF of $N(0,1)$

  • $\begingroup$ Can you explain a little more using p? $\endgroup$ – Japneet Singh Jun 4 at 17:19
  • 1
    $\begingroup$ In your Q, $p=\Phi(-1) = 1-\Phi(1)$. $\endgroup$ – Vizag Jun 4 at 17:19
  • $\begingroup$ Okay. Thanks for your help. $\endgroup$ – Japneet Singh Jun 4 at 17:35

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