How to Find Probability Density Function of a Continuous Random Variable

A function whose value at any given sample in the sample space can be explained as providing a relative likelihood that the value of the random variable would be equal to that sample is called a probability density function (PDF). It gives the probability that any value in a continuous set of values might occur. Its magnitude gives an idea of the likelihood of finding a continuous random variable near a certain point. A function defined on the outcomes of some probabilistic experiment which takes values in a continuous set is termed as a continuous random variable. These describe outcomes in probabilistic situations where the possible values some quantity can take form a continuum. In this article, we will discuss how to find the probability density function of a continuous random variable.

How To Find PDF of a Continuous Random Variable

We can find PDF using the formula P(aXb)=abfX(x)dxP(a\leq X\leq b) = \int_{a}^{b}f_{X}(x)dx

If the random variable can be any real number, then PDF is normalized such that fX(x)dx=1\int_{-\infty }^{\infty }f_{X}(x)dx = 1

The probability that X takes value between -∞ to +∞ is 1.


The non-normalized probability density function of a certain continuous random variable X is F(x) = 1/(1+x2). Find the probability that X is greater than 1, P(X>1)


The probability density function should be normalized.

11+x2dx=arctan(x)=π\int_{-\infty }^{\infty }\frac{1}{1+x^{2}}dx = arc \: tan(x)_{-\infty }^{\infty } = \pi

The normalized Probability density function is f~=1π(1+x2)\tilde{f} = \frac{1}{\pi (1+x^{2})}


P(X>1)=11π(1+x2)P(X>1) = \int_{1}^{\infty }\frac{1}{\pi (1+x^{2})} =1π arc tan(x)1= \frac{1}{\pi } \ arc \ tan (x)_{1}^{\infty }


= 1π(π2π4)\frac{1}{\pi} (\frac{\pi}{2}-\frac{\pi}{4})

= 1/4

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