Let the probability density function of a random variable $X$ be given by

$f(x)=\alpha e^{-x^2-\beta x}\ \ \ \ \ \ \infty<x<\infty $

If $E(X)= -\dfrac{1}{2}$ then

$(A) \ \alpha =\dfrac{e^{\frac{-1}{4}}}{\sqrt \pi}$ and $\beta=1$

$(B) \ \alpha =\dfrac{e^{\frac{-1}{4}}}{\sqrt \pi}$ and $\beta=-1$

$(C) \ \alpha =e^\frac{-1}{4}\sqrt \pi$ and $\beta=-1$

$(D) \ \alpha =e^\frac{-1}{4}\sqrt \pi$ and $\beta=1$

The way I did this question:

$\int_{-\infty}^{\infty}f(x)dx=\int_{-\infty}^{\infty}\alpha e^{-(x^2+\beta x +\frac{\beta^2}{4}-\frac{\beta^2}{4})}dx=\int_{-\infty}^{\infty}\alpha e^{-(x+\frac{\beta}{2})^2}e^{\frac{\beta^2}{4}}dx=1 $

put $y=(x+\frac{\beta}{2})$

$\int_{-\infty}^{\infty}\alpha e^{-(y)^2}e^{\frac{\beta^2}{4}}dx=\alpha\sqrt{\pi}e^{\frac{\beta^2}{4}}=1\implies \alpha=\dfrac{e^{\frac{-\beta^2}{4}}}{\sqrt{\pi}}$

Now using $E(X)=\int_{-\infty}^{\infty}x\alpha e^{-(x+\frac{\beta}{2})^2}e^{\frac{\beta^2}{4}}dx$

put $y=(x+\frac{\beta}{2})$

$E(X)=\int_{-\infty}^{\infty}(y-\frac{\beta}{2})\alpha e^{-(y)^2}e^{\frac{\beta^2}{4}}dx=\underbrace{\alpha e^{\frac{\beta^2}{4}} \int_{-\infty}^{\infty}(y) e^{-(y)^2}dx}_{\text {odd function}}-\frac{\beta}{2}\underbrace{\int_{-\infty}^{\infty}\alpha e^{-(y)^2}e^{\frac{-\beta^2}{4}}}_{\text{pdf}}dx$

$0-\dfrac{\beta}{2}=-\dfrac{1}{2} \implies \fbox {$\beta$ =1}$ and $\fbox{$\alpha=\dfrac{e^{\frac{-1}{4}}}{\sqrt{\pi}}$} $

Somehow after doing rigorous calculations, I got this answer but this question came in 2 marks so I am looking for a quick way out. Any alternate solution which less time consuming ?


$f(x)$ is in the form of a normal PDF: $\frac{1}{\sqrt{2\pi\sigma^2}}\exp(-(x-\mu)^2/2\sigma^2)$, which should be equal to $\alpha\exp(-x^2-\beta x)$. Coefficients should match in LHS and RHS, so $2\sigma^2=1$ and $E[X]=\mu=-1/2$. Then, LHS becomes $$\frac{1}{\sqrt{\pi}}\exp(-x^2+2\mu x - \mu^2)=\alpha\exp(-x^2-\beta x)$$ which gives you $\beta=1$ and $\alpha=\frac{\exp(-1/4)}{\sqrt{\pi}}$.

  • $\begingroup$ Omg I feel so dumb now . $\endgroup$ – Daman deep Jan 27 at 13:07
  • $\begingroup$ no you're path independent :) $\endgroup$ – gunes Jan 27 at 13:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.