Conditional PDF of Normal Distribution $X \sim N(0,1)$ , given $X>0$

conditional probabilityconditional-expectationnormal distributionprobability distributions

Given a Normal Random variable $X \sim N(0,1)$. I need to find the following:

  1. CDF, PDF of $X$, given $X>0$

  2. $E[X|X>0]$ and $Var(X|X>0)$

I have done the following for CDF:

$$P(X\le x|X>0) = \frac{P(0 \le X \le x)}{P(X>0)} = 2\Phi(x) – 1$$

And, hence conditional PDF will be : $$f_X(x|X>0) = 2.f_X(x), x \in [0, \infty)$$

Intuitively looking at Var(X|X>0), it will be same as Var(X), since Normal Distribution is a symmetrical one and the measure of variation around central tendency will be same even if $X>0$ because $Range_X$ has become half too.

Is this correct?

Best Answer

Since you have the conditional density, why not use it? You have

$$E[X|X>0] = \int_0^\infty xf_X(x|X>0) dx = 2\int_0^\infty f_X(x)dx,$$

and a similar formula for $\operatorname{Var}(X|X>0)$. You will find that the conditional variance will not be the same as the unconditional variance. To provide some intuition for this, we know that a standard Gaussian random variable lies in $(-1,1)$ with probability roughly $70\%$, so conditional on $X>0$, we should have $X\in(0,1)$ with probability roughly $70\%$. This suggests a smaller variance, which you can verify.

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